Tess & Kenton Chapter 6- Thermochemistry
Tess: Classify common processes as endothermic or exothermic
Endothermic reactions occur when the system absorbs energy from the surroundings; in other words, heat is supplied to the system by the surroundings
Melting, vaporization, and sublimation are common processes that are classified as endothermic
Example of endothermic reaction: energy + 2HgO(s) → 2Hg(l) + O2(g)
Exothermic reactions occur when the system releases energy into the surroundings; in other words, any process that gives off heat
Freezing and condensation are common processes that are classified as exothermic
Example of exothermic reaction: 2H2 (g) + O2(g) → 2H2O (l) + energy
Use thermochemical equations and stoichiometry to determine amount of heat lost or gained in a chemical reaction
Thermochemical equations are used to calculate the enthalpy( H, used to quantify the heat flow into or out of a system in a process that occurs at constant pressure) released or absorbed in a chemical reaction as well as the mass relationships
Rules for thermochemical equations:
1. Always specify the physical states of all reactants and products
Ex: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g) ∆H = -802.4 kJ/mol
2. What you do to the thermochemical equation, you must do to ∆H
Ex: H2O (s) → H2O (l) ∆H = 6.01 kJ/mol
2(H2O (s) → H2O (l)) ∆H = 2(6.01kJ/mol)
3. When we change the roles of the reactants and products (or reverse an equation), ∆H’s sign changes
Ex: H2O (s) → H2O (l) ∆H = 6.01kJ/mol
H2O (l) → H2O (s) ∆H = -6.01kJ/mol
Example: How much heat is evolved when 266 g of White phosphorus burn in air?
Thermochemical equation- P4 (s) + 5O2 (g) → P4O10 (s) ∆H = -3013kJ/mol
266g (P4) 1mol (P4) 3013 kJ 123.9g (P4) 1 mol (P4) Stoichiometry: 6471kJ Perform Calculations involving specific heat, mass, and temperature change
Heat Capacity (C)- amount of heat energy needed to raise temperature by 1° C
C=q/∆T [q=amount of heat ∆T=temperature change]
Specific Heat Capacity (c) or (s)- amount of energy needed to raise 1 gram of a substance by 1°C
q=mc∆T [q=amount of heat m=mass ∆T=temperature change]
Ex: The specific heat of graphite is .71 J/g°C. Calculate the energy needed to raise the temperature of 75kg of graphite from 294K to 348K.
q= (75000g) (.71J/g°C) (54K) = 2900kJ
Determine heats of reactions given experimental data collected in a calorimetry experiment
Calorimetry is the measurement of heat changes (a calorimeter is a closed, insulated container that usually measures the change in temperature of water but often the heat capacity is known
qrxn + qsoln = 0; qrxn = -qsoln if endothermic; qsoln = -qrxn if exothermic
Example: A 46.2 g sample of copper is heated to 95.4°C and then placed in a calorimeter containing 75.0 grams of water at 19.6 °C. The final temperature of both the water and the copper is 21.8°C. What is the specific heat of copper?
-qCu = qH2O
[copper] -mc∆T = mc∆T [water]
-[(46.2g)c(21.8-95.4)] = (75.0g)(4.184J/g°C)(21.8-19.6)
c = 0.203 J/g°C
Calculate standard enthalpy of reactions given the standard enthalpy of formations for products and reactants
Σ∆°Hf products - Σ∆°Hf reactants = Σ∆°Hreaction
Example: Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g) → 12CO2 (g) +6H2O (l)
∆°Hrxn= [12(-393.5) + 6(-285.8)] – [2(49.04)] = -6535kJ/2mol = -3267kJ/mol Apply Hess’s law to a multi-step process to determine standard enthalpy of reaction
Given: O2 (g) + H2 (g) → 2OH (g) ∆°H = +77.9kJ
O2 (g) → O2 (g) ∆°H = +495kJ
H2 (g) → 2H (g) ∆°H = +435.9kJ
Calculate the ∆°H for this reaction: O(g) + H(g) → OH(g)
[O2(g) + H2(g) → 2OH(g)]/2 ∆°H = [77.9]/2 [2O(g) → O2(g)]/2 ∆°H = [-495]/2
[2H(g) → H2(g)]/2 ∆°H = [-435.9]/2
38.95 + -247.5 + -217.95 = -426.5kJ
Problems:
1. A chemical reaction that absorbs heat from the surroundings is said to be and has a ∆H at constant pressure. A. endothermic; positive B. endothermic; negative C. exothermic; negative D. exothermic; positive E. exothermic; neutral (Answer: A because endothermic reactions take in energy from the surroundings making ∆H positive)
2. The specific heat capacity of lead is 0.13J/g-K. How much heat is required to raise the temperature of 15 g of lead from 22°C to 37°C? A. 29 B. 5.8e-4 C. -0.13 D. 2.0 E. 0.13 (Answer: A. because q=mc∆T and q=(15g)(0.13J/g-K)(15K)= 29.25)
Kenton:
Enthalpy Diagrams:
∆H is the change in enthalpy. If the products have less energy than the reactants, the ∆H is negative and the reaction is exothermic. If the products have more energy than the reactants, the ∆H is positive and the reaction is endothermic.
State Functions:
State Functions are a property of a system that depends only on the current state of the system. Examples include Enthalpy, Gibbs free energy, and Entropy. On the other hand, Non-State Functions or Path Functions are a property that depends on the path the system takes. Examples include work and heat.
First Law of Thermodynamics:
First Law of Thermodynamics: Energy can neither be created nor destroyed. The equation for the First Law of Thermodynamics is ∆E = q + w where ∆E stands for change in internal energy, q stands for heat, and w stands for work.
Sign Conventions for work and heat: When heat is absorbed by the system heat is positive, when heat is absorbed by the surroundings heat is negative. When work is done on the surroundings heat is negative, when work is done by the surroundings work is positive.
Gas-Phase Problems:
w=-P∆V is the equation for the work done by the gas on the surroundings.
Work and heat depend on the path a process takes making them Non-State Functions or Path Functions. Because of this, we cannot write ∆w because it does not only depend on the initial and final work, but it’s path to get there.
Internal Energy
∆U=q+w is the equation for change in internal energy.
Multiple Choice Questions:
In a certain process 50 J of work is done on the system and 30 J of heat are transferred to the surroundings. The change in internal energy of the system for this process, in J, is: a) 80 b) -80 c) 20 d) -20 e) 0
Answer: 20J because the 50J of work being done on the system based off of the sign convention for work is positive. The 30J of heat transferred to the surroundings is negative based of sign convention for heat and so the equation would be ∆U = -30J + 50J.
For a process taking place in an isolated system, which of the following must be true? a) internal energy is constant b) temperature is constant c) pressure is constant d) none
Answer: Internal energy is constant because no energy can be displaced somewhere else. Temperature and pressure can be manipulated to alter each other in an isolated system.
Important Pages:
Pg 233 – Figure 6.3 … Pg 236 Table 6.1 … Pg 237 Figure 6.5