Energetics/Thermochemistry

Energetics is the study of energy changes in chemical reactions. It is also called thermochemistry because these energy changes are usually seen as changes in temperature because the energy absorbed or evolved is in the form of heat.

It is based on the fact that energy is stored in chemicals when bonds are formed. The change in energy occurs when bonds are broken and reformed. The stronger the bonding, the lower ∆ H and more stable the substance.

The total energy of a system is called its enthalpy or heat content, which has the symbol H. While we do not know the absolute value of H, instead we can easily determine, by measurement the change in enthalpy or heat content which has the symbol ∆ H.

Reaction may be described as endothermic or exothermic depending on whether heat energy is absorbed or evolved. Complete the following table-:

Type of reaction Heat energy change Temperature change Sign of ∆ H

Exothermic

Endothermic

These two types of reactions may also be represented using energy diagrams:

H

products

∆H products

reactants ∆ H

reactants

In this case the reactants are more stable than the products as their energy level is lower. What type of reaction is this? Draw similar diagrams for the other type of reaction. What does the peak represent in the second diagram? How do catalysts and enzymes affect this peak? Show using an energy diagram.

The energy change in a reaction is represented by a thermochemical equation. This is an equation with an accompanying ∆ H value. This enthalpy change relates exactly to the molar amounts in the equation. e.g. if the coefficients in the equation are doubled, the ∆ H value will be doubled. 2Mg + O2 2MgO ∆ H = -1202 kJ mol-1 4Mg + 2O2 4MgO ∆ H = -2404 kJ mol-1 Question When liquid carbon disulfide, CS2, burns in excess oxygen, the equation for the reaction is: CS2(l) + 3O2(g) CO2(g) + 2SO2(g) ∆ H = -1072 kJ mol-1 a) What will be the heat evolved if the following amounts of CS2 are burnt in excess oxygen? i) 5.00 mol ii) 0.100 mol iii) 0.243 mol iv) 38.0g v) 3.5 kg b) What mass of CS2 will be needed to be burnt in excess oxygen to produce the following amounts of heat energy? i) 20 kJ ii) 5000 kJ iii) 685 J iv) 100 MJ Calculating Enthalpy Changes The amount of heat energy released or absorbed can be calculated using:

Heat energy = ∆ H = m . s . ∆ T where m = mass, ∆ T = change in temperature and s = specific heat capacity (the amount of heat required to raise the temperature of 1 gram of the sample by one degree - either Celsius or Kelvin) For water s = 4.182 J g-1 K-1, compared to alauminium which is 0.90, lead 0.13, iron 0.45 and ethanol 2.46. This formula can easily be used to determine ∆ H values for those reactions occurring in solution Molar heat of vaporisation - ∆ H for conversion of 1 mole from liquid to gas Molar heat of solution - ∆ H for 1 mole of solute to dissolve in water Molar heat of neutralisation - ∆ H for 1 mole of acid (or base) to be neutralised Heat of Combustion - ∆ H when a defined amount of substance reacts completely with oxygen

Question 1 A 100g heated lump of gold was dropped into 100 g of water, which was initially at 20.0˚ C. The final temperature of the mixture was 30.0˚ C. Assuming that the heat lost by the gold equals the heat gained by the water, what was the inital temperature of the gold? (s(Au) = 128 J kg-1 K-1 or 0.128 J g-1 K-1) Question 2 1.00 L of water is heated from 20.0˚ C to 50.0 ˚ C. Calculate the amount of heat required, given that the density of water is 1.00 g/cm3.

HESS’ LAW When ∆ H can’t be determined directly for an equation, it can be calculated by using more than equation which when combined produce the original equation/reaction. This is based on the premise that it doesn’t matter how many steps it takes reactants to get to the required products, the enrgy change will be the same. This is the basis of Hess’ Law. e.g if H2O(s) H2O(g) ∆H1 H2O(s) H2O(l) ∆H2 H2O(l) H2O(g) ∆H3 Then if equations 2 and 3 are added they will give the first equation. This means ∆ H1 = ∆ H2 + ∆ H3

This can also be shown in the form of an enthalpy cycle ∆H1 H2O(s) H2O(g)

∆ H2 ∆H3

H2O(l) If ∆ H2 = +6.00 kJ mol-1 and ∆ H3 = +44.01 kJ mol-1, what is ∆ H1?

If you have to reverse and equation, change the ∆ H sign to the opposite. e.g + to - Questions 1. The molar hat of sublimation of solid sulfur trioxide is 67.2 kJ mol-1 and its molar heat of fusion (s-l) is 20.3 kJ mol-1. Calculate the molar heat of vaporisation of liquid sulfur trioxide.

2. Find ∆ H for CO2(g) + H2(g) CO(g) + H2O(g) using H2(g) + 1/2O2(g) H2O(g) ∆H = -242 kJ mol-1 CO(g) + 1/2O2(g) CO2(g) ∆H = -282 kJ mol-1

3. Calculate ∆ H for the reaction 2S(s) + 3O2(g) 2SO3(g) given S(s) + O2(g) SO2(g) ∆H = -297.0 kJ mol-1 2SO2(g) + O2(g) 2SO3(g) ∆H = -196.0 kJ mol-1

4. Calculate ∆ H for the reaction C(s) + 2H2(g) CH4(g) given C(s) + O2(g) CO2(g) ∆H = -393 kJ mol-1 H2(g) + 1/2 O2(g) H2O(g) ∆H = -242 kJ mol-1 CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ∆H = -803 kJ mol-1