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GENERAL CHEMISTRY 2

nd 2 Semester - Module 2

THERMOCHEMI STRY

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General Chemistry 2 – G11/12 Support Material for Independent Learning Engagement (SMILE) Module 5: Thermochemistry First Edition, 2021

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What I Need to Know This module was designed and written to help you master the concepts of thermochemical equations and perform calculations involving thermochemistry.

The module is divided into three lessons, namely: Lesson 1 – First Law of Thermodynamics Lesson 2 – Enthalpy of the Reaction Lesson 3 – Hess Law

After going through this module, you are expected to: 1. Explain the First Law of Thermodynamics (STEM_GC11TCIIIg-i-124) 2. Explain Enthalpy of the Reaction (STEM_GC11TCIIIg-i-125); and 3. Calculate the change in enthalpy of a given reaction using Hess Law (STEM_GC11TCIIIg-i-127)

What's In

Activity 1: Forms of Energy Direction: Complete the table below by listing the form(s) of energy that is primarily generated from the given source. The first one is done for you.

ENERGY SOURCE FORM(S) OF ENERGY Wind mechanical Sun 1. 2. Geothermal 3 Fossil fuels 4. 5. Plants used as a food 6.

Activity 1 enabled you to review the different sources and forms of energy. Note that aside from the forms of energy identified, know that there are other forms of energy like sound, electrical, magnetic, and nuclear. Energy can be transformed from one form to another.

What's New

Activity 2. Energy Transformation Direction: Determine the possible energy transformation in each of the following:

1. ______4. ______8.______2. ______5. ______9. ______3. ______6. ______10. ______7. ______

The First Law of Thermodynamics

To understand and perform any sort of thermodynamic calculation, we must first understand the fundamental laws and concepts of thermodynamics.

Heat is the transfer of thermal energy between two bodies that are at different temperatures and is not equal to thermal energy.

Work is the force used to transfer energy between a system and its surroundings and is needed to create heat and thermal energy transfer.

Both work and heat allow systems to exchange energy. The relationship between the two concepts can be analyzed through the topic of thermodynamics, which is the scientific study of the interaction of heat and other types of energy.

What is it

The First Law of Thermodynamics states that "energy cannot be created nor destroyed. It can be transformed into another form, but the total amount of energy remains the same. How energy is conserved is shown when you eat your meal. The chemical energy in the food will be converted into mechanical energy that enables you to perform your daily task. But not all the chemical energy from the food you take will be transformed into mechanical energy. Some of it will be released from your body as heat when you sweat or feel warm. The law deals with energy, work, and heat.

According to the Law of Conservation of Energy, energy can neither be created nor destroyed. It can only be transferred between the system and the surroundings. The energy of the system (ΔEsys) and the energy of its surroundings (ΔEsurr) are related by the following equation:

ΔEsys = –ΔEsurr

The negative sign indicates the flow of energy. As the system releases energy, the surroundings absorb it. Whatever amount of energy that is lost by the surroundings must be gained by the system.

The First Law of Thermodynamics states that in the process, the change in energy of a system is equal to the heat absorbed (q) by the system and the work (w) done on it. ΔEsys = q + w Where: ΔEsys is the total change in internal energy of a system q is the heat exchanged between a system and its surroundings w is the work done by or on the system.

The law implies that the net energy flow to or from any system comes in the form of either work or heat. When work is done on the system, it gains energy, and work is denoted as positive (+w). When the system does the work, it uses up or transfers some of its energy so that work is denoted as negative (– w). Moreover, the system may also absorb heat (endothermic) for which q is positive (+q), or it may release heat (exothermic) to the surroundings for which q is negative (–q). Thus, the net change in energy can either be positive or negative, depending on the means and direction of the energy transfer involved.

The table below shows the sign of the values of the q and w as it enters or leaves the system.

Heat added to the system → → Heat evolved by the system

(+q) SYSTEM (-q) Endothermic Exothermic

Work done on the system → → Work done by the system (+w) (–w)

Heat (q) and work (w) are positive when they enter the system and negative when they leave the system. As to the internal energy (ΔEsys), the table preceding table serves as a guide in determining the sign of the value of ΔEsys.

q w ΔEsys (internal energy)

+ + + (internal energy increases)

– – – (internal energy decreases)

+ – The sign of ΔE depends on the – + magnitude of q and w.

The internal energy of a system would decrease if the system gives off heat or does work. Therefore, the internal energy of a system increases when the heat increases (this would be done by adding heat into a system). The internal energy would also increase if work were done on a system. Any work or heat that goes into or out of a system changes the internal energy. However, since energy is never created nor destroyed as the first law of thermodynamics states, then the change in internal energy always equals zero. If energy is lost by the system, then it is absorbed by the surroundings. If energy is absorbed into a system, then that energy was released by the surroundings.

Sample Problem

A gas in a system has constant pressure. The surroundings around the system lose 62 J of heat and do 474 J of work onto the system. What is the internal energy of the system?

Solution

To find internal energy, ΔEsys, we must consider the relationship between the system and the surroundings. Since the First Law of Thermodynamics states that energy is not created nor destroyed. We know that anything lost by the surroundings is gained by the system. The surrounding area loses heat and does work on the system.

ΔEsys = q + w ΔEsys = (62J) + (474J) = 536J (Endothermic)

Activity 3: Calculate the change in energy for the following processes. Determine if the process is endothermic or exothermic.

1. A gas releases 35 J of heat as 84 J of w was done to compress it. ______

2. A gas absorbs 48 J of heat as it does 72 J of work by expanding. ______

An interesting observation whenever energy is utilized to do work is that some of it seem to escape. It just doesn't go to where we want it to go. That means that the utilization of energy is almost always less than 100% efficient. If the released energy is not used, it will escape and be "wasted," usually as heat.

All forms of energy can be quantitatively converted into heat. For example, when a moving car stops, where does all the mechanical energy go? It is converted entirely into heat by the frictional action of the brakes, which become very hot. When electrical energy is directed to a material that is a poor conductor of the electricity, such as a filament in flat irons, the electrical energy is converted entirely into heat. The full convertibility of the different forms of energy into heat gives us a way of measuring energy.

Activity 4: The Energy Deal Directions: A. Write TRUE if the statement is true and FALSE if the statement is not true.

______1. The heat added to a system is used to decrease the system’s internal energy. ______2. The First Law of Thermodynamics describes how chaotic is the energy of the universe. ______3. Steam engines operate by boiling water to produce steam. The work done by the steam is due to the expansion of water as it turns into steam. ______4. Heat is the transfer of thermal energy between two bodies that are of equal temperatures. ______5. Work is required to create heat and causes the transfer of energy between the system and its surroundings.

B. Refer to the First Law of Thermodynamics. Decide whether the statement below is TRUE or FALSE. Support your answer.

"Our energy supply is running out."

______Enthalpy of a Chemical Reaction

Most chemical reactions in laboratories and even in living systems occur under constant pressure. When such chemical reactions are carried out in vessels exposed to atmospheric pressure, energy changes between the system and the surroundings usually involve heat transfer only since the amount of work involved is negligible under this condition.

We have stated that the change in energy (ΔE) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called pressure-volume work or PV work. 5

Consider the example of work performed by a reaction carried out at constant pressure in a closed container, as shown in the figure below.

Initially, the system (a copper penny and concentrated nitric acid) is at atmospheric pressure, as shown in (a). When the penny is added to the nitric acid, the volume of NO2 gas that is formed causes the piston to move upward to maintain the system at atmospheric pressure, as shown in (b). In doing so, the system is performing work on its surroundings.

The chemical equation for this reaction is,

Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2H2O(l) + 2NO2

If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston rises as nitrogen dioxide gas is formed.

The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by the movement of the piston (ΔV). At a constant external pressure (here, atmospheric pressure), W = −PΔV.

The negative sign associated with PV work done indicates that the system loses energy when the volume increases. If the volume increases at constant pressure (ΔV>0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (ΔV<0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.

The internal energy ΔE of a system is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy (H) (from the Greek enthalpein, meaning "to warm").

The enthalpy of a system is defined as the sum of its internal energy ΔE plus the product of its pressure P and volume V:

H= ΔE + PV 6

Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. So, we can define a change in enthalpy (ΔH) accordingly by the equation:

ΔH = Hfinal −Hinitial

If a chemical change occurs at constant pressure (i.e., for a given P, ΔP = 0), the change in enthalpy (ΔH) is ΔH = Δ(E+PV) = ΔE + ΔPV = ΔE + PΔV

Substituting q +w for ΔE (First Law of Thermodynamics) and −w for PΔV, we obtain, ΔH = ΔE + PΔV = qp + w − w ΔH = qp

Where: ΔH = change in enthalpy qp = heat at constant pressure

From the equation above, we see that at constant pressure, the change in enthalpy, ΔH of the system, is equal to the heat gained or lost.

ΔH = Hfinal − Hinitial = qp

Just as with ΔE, because enthalpy is a state function, the magnitude of ΔH depends only on the initial and final states of the system, not on the path taken. Most importantly, the enthalpy change is the same even if the process does not occur at constant pressure.

When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction (ΔHrxn), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so ΔHrxn is negative.

Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so ΔHrxn is positive. Thus:

ΔHrxn <0 for an exothermic reaction, and ΔHrxn >0 for an endothermic reaction.

The sign conventions for heat flow and enthalpy changes are summarized in the following table:

Reaction Type Q ΔHrxn Exothermic < 0 – (heat flows from a system to its surroundings) Endothermic > 0 + (heat flows from the surroundings to a system)

If ΔHrxn is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill. Conversely, if ΔHrxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill.

Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.

In chemical reactions, bond breaking requires an input of energy and is, therefore, an endothermic process, whereas bond making releases energy, which is an exothermic process.

The Enthalpy of Reaction

Energy changes in chemical reactions are usually measured as changes in enthalpy. (a) If heat flows from a system to its surroundings, the enthalpy of the system decreases, ΔHrxn is negative, and the reaction is exothermic; it is energetically downhill. (b) Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, ΔHrxn is positive, and the reaction is endothermic; it is energetically uphill.

Activity 5: Endothermic or Exothermic Direction: Identify whether each process involves an exothermic or endothermic reaction. ______1. Melting ice ______2. Subliming naphthalene balls ______3. Thermal decomposition ______4. Dissolving ammounium chloride in water ______5. Photosynthesis ______6. Cellular respiration ______7. Water freezing into ice cubes ______8. Rusting of iron ______9. Burning of candleas ______10. Setting of cement and concrete

Enthalpy of Formation and Reaction

Standard States and Standard Enthalpy Changes

The standard enthalpy of formation refers to the enthalpy change when one mole of a compound is formed from its elements.

Standard States

In Chemistry, the standard state of a material, be it a pure substance, mixture, or solution, is a reference point used to calculate its properties under different conditions. In principle, the choice of the standard state is arbitrary, although the International Union of Pure and Applied Chemistry (IUPAC) recommends a conventional set of standard states for general use. A standard pressure of 1 bar (101.3 kilopascals) has been accepted.

Strictly speaking, temperature is not part of the definition of a standard state. The standard state of a gas is conventionally chosen to be 1 bar for an ideal gas, regardless of the temperature. However, most tables of thermodynamic quantities are compiled at specific temperatures, most commonly 298.15 K (exactly 25°C) or, somewhat less commonly, 273.15 K (exactly 0°C).

Standard states for atomic elements are given in terms of the most stable allotrope for each element. For example, white tin and graphite are the most stable allotropes of tin and carbon, respectively. Therefore, they are used as standard states or reference points for calculations of different thermodynamic properties of these elements.

Standard Molar Enthalpy of Formation, ΔHf

The standard enthalpy of formation, or standard heat of formation, of a compound, is the change in enthalpy that accompanies the formation of one mole of the compound from its elements in their standard states. For example, the standard enthalpy of formation for carbon dioxide would be the change in enthalpy for the following reaction:

Note that standard enthalpies of formation are always given in units of kJ/mol of the compound formed.

Standard Molar Enthalpy of Reaction, ΔHrxn

The standard molar enthalpy of formation can be calculated by subtracting the sum (∑) of the enthalpy of formation of the product and the sum of the enthalpies of formation of the reactants.

Consider the hypothetical equation, aA + bB → cC + dD

Enthalpy of formation of product: ∑ΔHf (products) = cΔHf(C) + ΔHf(D)

Enthalpy of formation of reactant: ∑ ΔHf (reactants) = aΔHf(A) + bΔHf(B)

Then the standard molar enthalpy of the hypothetical reaction is calculated as,

∑ ΔHf (rxn) = ΔHf (products) – ΔHf (reactants)

Sample Problem:

Consider the combustion of ethanol, C2H5OH to produce CO2 and H2O.

C2H5OH (g) + 3O2(g) → 2CO2(g) + 3H2O(l)

ΔHf (C2H5OH) = – 277.7 kJ ΔHf (O2) = 0 ΔHf (CO2) = – 393.5 kJ/mol ΔHf (H2O) = – 241.8 kJ/mol

∑ΔHf (rxn) = 2[ΔHf (CO2,g) + 3[ΔHf (H2O,g)]–[(ΔHf (C2H5OH,l)+3[ΔHf (O2)]

= 2[ΔHf (-393.5 kJ)+3[ ΔHf (-241.8)]–[(ΔHf (-277.7kJ)]+3[ΔHf (0)]

= [2(-3973.5 kJ/mol)]+[3(-231.8 kJ/mol)]–[-77.77kJ/mol+3(0)]

= [(-787 kJ) + (-695.4 kJ)] – [(-277.7 kJ) + (0)]

= [1482.4] – [-277.7 kJ]

= 1,204.7 kJ

Calculating the Standard Enthalpy of Reaction

Calculate the standard enthalpy of reaction for the combustion of methane:

To calculate the standard enthalpy of reaction, we need to look up the standard enthalpies of formation for each of the reactants and products involved in the reaction. These are typically found in an appendix or in various tables online. For this reaction, we need the ΔHf of the following substances: ΔHf{CH4(g)} = −74.84 kJ/mol ΔHf{CO2(g)} = −393.5 kJ/mol ΔHf{H2O(g)} = −241.8 kJ/mol ΔHf{O2(g)} = 0 kJ/mol because O2(g) is an element in standard state

Note that because it exists in its standard state, the standard enthalpy of formation for oxygen gas is 0 kJ/mol. Next, we sum up our standard enthalpies of formation. Keep in mind that because the units are in kJ/mol, we need to multiply by the stoichiometric coefficients in the balanced reaction equation.

∑ΔHf {products} = 1[ΔHf (CO2,g)] + [2(ΔHf (H2O,g)]

=[(1)(−393.5kJ/mol)] + [(2)(−241.8kJ/mol)]

= [−393.5kJ/mol] + [−483.6kJ/mol]

= −877.1 kJ/mol

∑ΔHf{reactants} = [ΔHf (CH4,g] + [2 ΔHf (O2,g)]

= [1(−74.84 kJ/mol] + [2(0)]

= −74.84 kJ/mol

Now, we can find the standard enthalpy of reaction,

ΔHrxn= ∑ΔHf (products) − ∑ΔHf (reactants)

= (−877.1 kJ/mol) − (−74.84 kJ/mol)

= −802.26 kJ/mol

As we would expect, the standard enthalpy for this combustion reaction is strongly exothermic.

Activity 6: Check Your Knowledge Temperature Direction: Calculate the heat of reaction of the combustion of ethane gas, C2H4. The balanced equation for the reaction is,

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)

ΔHf values: CO2(g) = −393.5 kJ/mol H2O(l) = −393.5 kJ/mol C2H4(g) = 53.3 kJ/mol O2(g) = 0

The enthalpy or enthalpy change is how much energy (in the form of heat) has been transferred out or taken in during a chemical reaction. Many compounds cannot be directly synthesized from their elements. In some cases, the reaction proceeds too slowly, or side reactions produce substances other than the desired compound. In these cases, ∆H°f (enthalpy change) can be determined by an indirect approach, which is based on Hess's Law of Heat Summation, or simply Hess's Law, named after the Swiss chemist, Germain Henri Hess.

Hess's Law

Hess's Law sums the changes in enthalpy for a series of intermediate reaction steps to find the overall change in enthalpy for a reaction.

This law states that if a reaction takes place in several steps, then the standard reaction enthalpy for the overall reaction is equal to the sum of the standard enthalpies of the intermediate reaction steps, assuming each step takes place at the same temperature.

Hess's Law derives directly from the Law of Conservation of Energy, as well as its expression in the First Law of Thermodynamics. Since enthalpy is a state function, the change in enthalpy between products and reactants in a chemical system is independent of the pathway taken from the initial to the final state of the system. Hess's Law can be used to determine the overall energy required for a chemical reaction, especially when the reaction can be divided into several intermediate steps that are individually easier to characterize.

A negative enthalpy change for a reaction indicates an exothermic process, while a positive enthalpy change corresponds to an endothermic process.

Activity 7: The Truth About Enthalpy Direction: Write TRUE if the statement is correct and FALSE if it is not.

______1. The enthalpy change accompanying a chemical change is independent of the route by which the chemical change occurs. ______2. Hess's Law says that if you convert reactants A into products B, the overall enthalpy change will be exactly different whether you do it in one step or two steps or however many steps. ______3. The enthalpy change is the heat evolved or absorbed during a reaction happening at constant pressure. ______4. All steps in enthalpy change have to proceed at the same temperature and the equations for the individual steps must balance out. ______5. Enthalpy is a state function that the magnitude of ΔH does not depend on the path.

Calculating Standard Enthalpies of Reaction Using Hess's Law

Problem 1. What is the ΔHrxn of the following reaction?

C(s){graphite} → C(s){diamond} ΔHrxn=?

Turning graphite into diamond requires extremely high temperatures and pressures, and therefore is impractical in a laboratory setting. The change in enthalpy for this reaction cannot be determined experimentally. However, because we know the standard enthalpy change for the oxidation of these two substances, we can calculate the enthalpy change for this reaction using Hess's Law.

Our intermediate steps are as follows:

C(s){graphite}+O2(g)→CO2(g) ΔH = −393.41 kJ/mol

C(s){diamond}+O2(g)→CO2(g) ΔH = −395.40 kJ/mol

In order to get these intermediate reactions to add to our overall net reaction, we need to reverse the second step. Keep in mind that when reversing reactions using Hess's Law, the sign of ΔH will change. Sometimes, you will need to multiply a given reaction intermediate by an integer. In such cases, you need always multiply your ΔH value by that same integer. Restating the first equation and flipping the second equation, we have:

C(s){graphite} + O2(g) → CO2(g) ΔH =−393.41 kJ/mol

CO2(g) → C(s){diamond} + O2(g) ΔH =+395.40 kJ/mol

Adding these equations together, carbon dioxides and oxygens cancel, leaving us only with C(s){graphite} and C(s){diamond} in our net equation. By Hess's Law, we can sum the ΔH values for these intermediate reactions to get our final value, ΔHrxn.

C(s){graphite} + O2(g) → CO2(g) ΔH=−393.41 kJ/mol

CO2(g) → C(s){diamond} + O2(g) ΔH =+395.40 kJ/mol

C(s){graphite} → C(s){diamond} ΔHrxn=1.99 kJ/mol

Problem 2: What is the value of ΔH for the following reaction?

Given: 2CO(g) + O2(g) → 2 CO2(g) Eq.1

2C(s) + O2(g) →2CO(g); ΔHf = -221.0 kJ/mol Eq. 2

C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol Eq. 3

Solution

Hess's Law says the total enthalpy change does not rely on the path taken from beginning to end. Enthalpy can be calculated in one grand step or multiple smaller steps.

Step 1: Manipulate given equations to most closely resemble the equation of interest. To solve this type of problem, organize the given chemical reactions where the total effect yields the reaction needed. There are a few rules that you must follow when manipulating a reaction. 1. The reaction can be reversed. This will change the sign of ΔHf. 2. The reaction can be multiplied by a constant. The value of ΔHf must be multiplied by the same constant.

 We see a resemblance of value from the equation of interest (Eq.1) to Equation 2

2CO(g) + O2(g) → 2 CO2(g) Eq 1

2C(s) + O2(g) →2CO(g) ; ΔHf = -221.0 kJ/mol Eq 2

 Flip Equation 2 so that 2CO(g) will be on the reactant side as reflected by the equation of interest. Applying rule #1, this will become:

2 CO(g) → 2C(s) + O2(g) ; ΔHf = +221.0 kJ/mol Eq 4

 Another resemblance from the equation of interest (Eq. 1) to equation 3. 2CO(g) + O2(g) → 2 CO2(g) Eq 1

C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol Eq. 3

 Both resemblances are on the product side, but their coefficients are not the same; therefore, multiply the whole equation 3 to have a coefficient of 2. Applying rule #2, this will become:

2C(s) + 2O2(g) → 2CO2(g); ΔHf = -787 kJ/mol Eq. 5

Step 2: Add new reactions together.  Add equation 4 and equation 5 together with their ΔHf This will become

Step 3: Cancel out any compounds that are the same on both sides of the reaction arrow. Anything that is the same on both of the reaction arrows can be canceled out

Step 4: Write the new equation and double-check to make sure it matches the equation of interest.

2CO(g) + O2(g) → 2CO2(g) : ΔHf = -566 kJ/mol

Therefore the value of ΔH for the following reaction,

2CO(g) + O2(g) → 2 CO2(g) is –566 kJ/mol

A chemical equation that shows the value and direction of heat involved in a reaction is specifically called thermochemical reaction. It also indicates the physical state of the reactants and products involved in the reaction.

The amount of heat absorbed or released by a chemical reaction at constant atmospheric pressure is called enthalpy (H). It is a form of chemical energy. The difference between the enthalpy of the products and the reactants is called enthalpy of a reaction (ΔH).

ΔH = Hproducts – Hreactants

When (ΔH) is positive, the chemical reaction is described as endothermic. When (ΔH) is negative, the reaction is exothermic. Consider the exothermic combustion of methane (CH4) into carbon dioxide (CO2) and water (H2O).

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -890.4 kJ/mol

This thermochemical equation means that 890.4 kJ of heat is released when one mole of gaseous methane and two moles of oxygen gas combust to form one mole of gaseous carbon dioxide and two moles of liquid water.

Enthalpy is an extensive property. Its magnitude is proportional to the amount of reactants and products in the reaction. For example, if the amount of reactants and products in a thermochemical equation is multiplied by 2, the value of the enthalpy is also multiplied by 2. If multiplied by ½ the enthalpy is also reduced to half.

Original equation: CH4(g) + 2O2(g) → CO2(g) + 2 H2O(l) ΔH = -890.4 kJ/mol

Multiplied by 2 : 2CH4(g) + 4O2(g) → 2CO2(g) + 4H2O(l) ΔH = -1780.8 kJ/mol

Multiplied by ½ : 1/2CH4(g) + O2(g) →1/2CO2(g) + H2O(l) ΔH = -445.2 kJ/mol

Activity 8: The Heat is On! Direction: Determine the ΔHrxn from ΔHf values

The reaction of liquid hydrogen peroxide, H2O2, with liquid hydrazine has also been suggested as a source of energy for small rockets. Find out ΔHrxn for this reaction and compare the usefulness of the reaction.

2H2O2(l) + N2H4(l) → N2(g) + 4H2O(g)

ΔHf values: H2O2(l) = –187.8 kJ/mol N2H4(l) = 50.6 jJ/mol N2(g) = 0 H2O(g) = –238.8 kJ/mol

What's More

Activity 9: Some More Heat Please Direction: Solve the following problems and show the complete solution. You may use a separate sheet of paper for your answer.

1. Calculate the ΔHrxn for the reaction,

Sn (s) + Cl2 (g) → SnCl2 (s) ΔHrxn = ?

Sn (s) + Cl2 (g) → SnCl2 (s) Eq.1

Sn (s) + 2Cl2 (g) → SnCl4 (l); ΔH = – 545.2 kJ Eq.2

SnCl2 (s) + vCl2 (g) → SnCl4 (l); ΔH = – 195.4 kJ Eq.3

2. Calculate for the enthalpy of the following reaction between dinitrogen tetroxide, N2O4, and liquid hydrazine, N2H4, used to fuel small rockets. The balanced equation for the reaction is,

N2O4(g) + 2 N2H4(l) → 3N2(g) + 4H2)(g)

ΔHf values: N2O4 (g) = +9.2 kJ/mol H2O(g) = −238.9 kJ/mol N2H4(g) = 50.6 kJ/mol N2(g) = 0

What I Have Learned The Choice Is Yours I. Select and write the letter that corresponds to the best answer in the space provided. _____1. Changes in enthalpy of an exothermic reaction is A. Positive C. negative B. Constant D. neutral

_____2. The First Law of Thermodynamics states that energy can not be A. created only C. destroyed only B. converted D. created and destroyed

_____3. Hess's Law states that a chemical reaction is independent of the route by which chemical reactions takes place while keeping the same A. initial and final conditions C. initial conditions only B. final conditions only D. mid-conditions

_____4. The change in the energy between a chemical reaction and the surroundings at constant temperature is called A. enthalpy C. enthalpy profile B. enthalpy change D. dynamic enthalpy

_____ 5. The application of the law of thermodynamics to the enthalpy change was proposed by A. Newton C. Hess’s B. Lewis D. Sophocles

II. Choose the letter of the option that completes the concept.

6. The thermochemical equation showing the formation of ammonia, NH3 from its elements is: N2 (g) + 3 H2 (g) —> 2 NH3 (g); ΔH = - 92 kJ. This equation shows that 92 kJ of heat is ______. A. lost to the surroundings B. gained in the surroundings

7. At constant pressure, enthalpy is? ______. A. equal to the enthalpy change of the system. B. equal to the total energy of the surroundings

8. A reaction is allowed to take place in an insulated container containing 100 mL of water. If the reaction is exothermic, what happens to the temperature of the water? ______. A. The temperature of the water goes up B. The temperature of the water goes down

9. If ΔH is positive, then it is? ______. A. endothermic B. exothermic

10. If heat is released by a chemical system, an equal amount of heat will be? A. Absorbed by the system B. Absorbed by the surroundings

Assessment

Direction: Encircle the letter of the best answer.

1. The First Law of Thermodynamics is the Law of ______A. Conservation of energy B. Conservation of work C. Conservation of mass D. Conservation of heat

2. The First Law of Thermodynamics is about ______. A. entropy C. temperature B. enthalpy D. internal energy

3. Which statement about enthalpy is true? A. Heat is given off to the surroundings in endothermic reactions. B. The sign of ∆H is always negative in exothermic reactions. C. Some substances have a negative specific heat capacity. D. Specific heat capacity is the same for all liquids.

4. What happens to the value of ∆H for a thermochemical reaction if the reaction is reversed? A. ∆H is the reciprocal of the original value, and the sign remains the same. B. ∆H has the same numerical value, and the sign remains the same. C. ∆H is the reciprocal of the original value and the sign changes. D. ∆H has the same numerical value and the sign changes.

5. Which is an exothermic process? A. Water vapor condensing C. Water boiling B. Water evaporating D. Ice melting

6. Which is not true about entropy? A. It is denoted by the symbol S B. It is a measure of the randomness or disorder of a system. C. It is a measure of how much energy is unavailable for conversion into work. D. A positive value of a change in entropy indicates that the final state is less random than the initial state.

7. Which reaction below is endothermic? A. 2NO2(g) ⇌ N2O4(g), ∆H = -58.0 kJ B. PCl5(g) ⇌ PCl3(g) + Cl2(g), ∆H = 87.9 kJ C. N2(g) ⇌ 3H2(g) + 2NH3(g), ∆H = -92.38 kJ D. 2PCl3(g) + O2(g) ⇌ 2POCl3(g), ∆H = -508 kJ

8. The branch of thermodynamics which deals with the interconversion of energy between heat and work. A. thermochemistry B. Kinetics C. thermodynamics D. equilibria

9. Which of the following equations represents an endothermic reaction? A. 2H2(g) + O2(g) → 2H2O(l) ΔH = -572 kJ B. 2H2(g) + O2(g) → 2H2O(l) + 572 kJ C. 2BrCl(g) + -29.3 kJ → Br2(g) + Cl2(g) D. N2O4(g) + 59 kJ → 2NO2(g)

10. Which of the following is true following the law of conservation of energy? A. ΔEsys + ΔEsurr = 0 B. ΔEsys = ΔEsurr C. ΔEsys = –ΔEsurr D. ΔEsys – ΔEsurr = 0

11. Which of the following will have a standard enthalpy of formation equal to zero? A. N2 (g) B. CO2 (g) C. NO2 (g) D. MgCl2 (g)

12. Living plants produce glucose during photosynthesis according to the equation: 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) Is the reaction endothermic or exothermic, and is the value of ΔH positive or negative? A. exothermic, positive B. exothermic, negative C. endothermic, positive D. endothermic, negative

13. Given the hypothetical thermochemical equation: A + B → C + D ΔH = - 430 kJ. Which among the following statements is correct about this reaction? A. The reaction is endothermic. B. The equation may be written as A + B + 430 kJ → C + D C. The heat content of A and B is greater than the heat content of C and D D. The heat content of C and D is greater than the heat content of A and B.

° 14. Given the following data: 2 O3(g) → 3 O2(g) ΔH = -427 kJ ° O2(g) → 2 O(g) ΔH = +495 kJ ° NO(g) + O3(g) → NO2(g) + O2(g) ΔH = -199 kJ Which of the following option shows the ΔH° for the reaction, NO(g) + O(g) → NO2(g)? A. 233 kJ C. -465 kJ B -233 kJ D. 465 kJ

15. Given: CH3OH(l); ΔH = –238.6 kJ/mol CO2(g) ; ΔH = –393.5kJ/mol H2O (l) ; ΔH = –285.8kJ/mol The standard enthalpy change, ΔH°, for the following reaction,

2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O (l) A. ΔH = –1299.5 kJ B. ΔH = –1075 kJ C. ΔH = –424.8 kJ D. ΔH = –3267.8 kJ

Additional Activity Reflection:

I learned that ______

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I enjoyed most on______

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I want to learn more on ______

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Answer Key General Chemistry 2 Module 5

What’s In: Activity 1 What Is It 1. Light, Activity 6: ΔHrxn = 1410.9 kJ/mol 2. heat 3. heat Activity 7: 4. light 1. True 5. heat 2. False 6. chemical 3. True 4. True What’s New: Activity 2 5. True 1. mechanical 2. electrical Activity 8: 3. light 2H2C2 + N2H4(l) → N2(g) + 4H2O(g) 4. electrical ΔHrxn = –630.6 kJ/mol 5. light 6. heat What’s More 7. sound 1. ΔHrxn = 349.8 kJ/mol 2. ΔHrxn = 1066 kJ/mol 8. chemical 3. 60.025 g 9. mechanical 4. 1,740 L 10. sound What I Have Learned What Is It 1. C Activity 3: 2. D 1. 49 J – Endothermic 3. A 2. –24 J – Exothermic 4. B 5. C Activity 4: 6. A 1. 1. False 7. A

2. False 8. B 3. True 9. A 4. False 10. B 5. True 2. Not True

Activity 5: 1. Endothermic Assessment 2. Endothermic

3. Endothermic 1. A 6. A 11. A 4. Endothermic 2. D 7. B 12. C 5. Endothermic 3. B 8. A 13. D 6. Exothermic 4. D 9. D 14. B 7. Exothermic 5. A 10. B 15. A 8. Exothermic 9. Exothermic 10. Exothermic

References

Ayson, M, et al. General Chemistry 2 Senior High School Textbook, 2016

Bayquen, A and Gardee Peña. General Chemistry 2. Phoenix Publishing House, 2016.

Fajardo, N and Macario Catahan. Chem C8 Chemistry and the Environment, UPOU Learning Resource, 1994

Rodriquez, M, and Ma. Cecilia de Mesa. Fundamental Concepts of Chemistry II, UPOU Learning Resource, 1994

Online Sources

Enthalpy of Reaction accessed from https://chem.libretexts.org/Bookshelves/ General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05._ Thermochemistry/5.4%3A_Enthalpy_of_Reaction#:~:text=For%20a%20chemical%20rea ction%2C%20the,%CE%94Hrxn.

Image of a television accessed at https://www.moving.com/tips/how-to-ship-a-tv/ on February 8, 2021

Image of a trumpet accessed at https://www.dreamstime.com/stock-photo-man-woman- playing-trumpet-white-player-black-room-saxophone-people-female-adult-male-play- two-couple-music-instrument-image97710546 on February 8, 2021

Image of turbine accessed at https://www.bikersrights.com/wind-turbine-for-a-school- project/ on February 8, 2021

Thermochemistry Lumen Boundless Chemistry: https://courses.lumenlearning.com/ boundless-chemistry/chapter/ standard-enthalpy-of-formation-and-reaction/#:~:text= The%20standard%20enthalpy%20of%20formation%20is%20the%20change%20in%20e nthalpy, transformed%20by%20a%20chemical%20reaction.