Hess's Law: 

The total amount of heat evolved or absorbed in a reaction is the same whether the reaction takes place in one step or in a no. of steps.

Relationship between Cp and Cv

Intext: - 

Problem 6.1 Express the change in internal energy of a system when 

(i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. What type of wall does the system have ? 

(ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have? 

(iii) w amount of work is done by the system and q amount of heat is supplied to the system. What type of system would it be? 

Solution: 

(i) ∆ U = w ad, wall is adiabatic 

(ii) ∆ U = – q, thermally conducting walls

(iii) ∆ U = q – w, closed system. 

Problem 6.2 Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ? 

Solution We have q = – w = p (10 – 2) = 0(8) = 0 No work is done; no heat is absorbed. 

Problem 6.3 Consider the same expansion, but this time against a constant external pressure of 1 atm. 

Solution We have q = – w = pex (8) = 8 litre-atm 

Problem 6.4 Consider the expansion given in problem 6.2, for 1 mol of an ideal gas conducted reversibly. 

Solution We have q = – w = 2.303 nRT log Vf/ Vs = 2.303 × 1 × 0.8206 × 298 × log 10 /2 

= 2.303 x 0.8206 x 298 x log 5

= 2.303 x 0.8206 x 298 x 0.6990 

= 393.66 L atm 

Problem 6.5 If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41kJ mol–1. Calculate the internal energy change, when 1 mol of water is vapourised at 1 bar pressure and 100°C. 

Solution (i) The change H2O (l) → H2O(g) 

∆H = ∆U + ∆ngRT

∆U = ∆H – ∆ngRT,

substituting the values, we get 

∆U =41.00 kJ/mol - 1x8.3 J / mol K x 373 K

= 41.00 kJ / mol - 3.096 kJ / mol 

= 37.904 kJ / mol 

Problem 6.6 1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C (graphite) + O2 (g) → CO2 (g) During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm? 

Solution Suppose q is the quantity of heat from the reaction mixture and CV is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter. 

q = Cv × ∆T 

Quantity of heat from the reaction will have the same magnitude but opposite sign because the heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter. 

q=-Cv x ∆T = -20.7 kJ/ K x (299-298) K = -20.7kJ 

(Here, negative sign indicates the exothermic nature of the reaction).

Thus, ∆U for the combustion of the 1g of graphite = – 20.7 kJK–1 

For combustion of 1 mol of graphite, 

= 12.0 g/mol x (-20.7 kJ) / 1g = -2.48x 102 kJ/mol 

since ∆ng = 0

∆H = ∆U = -2.48x 102 kJ/mol

Problem 6.7 A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporisation at 298K. ∆vapH  for water at 298K= 44.01kJ mol–1 

Problem 6.8 Assuming the water vapour to be a perfect gas, calculate the internal energy change when 1 mol of water at 100°C and 1 bar pressure is converted to ice at 0°C. Given the enthalpy of fusion of ice is 6.00 kJ mol1 heat capacity of water is 4.2 J/g°C 

Ans. The change take place as follows: 

Step - 1 1 mol H2O (l, 100°C) à 1 mol (l, 0°C) Enthalpy change ∆H1 

Step - 2 1 mol H2O (l, 0°C) à 1 mol H2O( S, 0°C) Enthalpy change ∆H2 

Total enthalpy change will be - ∆H = ∆H1 + ∆H2 

∆H1 = - (18 x 4.2 x 100) J mol-1 = - 7560 J mol-1 = - 7.56 k J mol-1 

∆H2 = - 6.00 kJ mol-1 

Therefore, 

∆H = - 7.56 kJ mol-1 + (-6.00 kJ mol-1) = -13.56 kJ mol-1 

There is negligible change in the volume during the change form liquid to solid state. 

Therefore, p∆v = ∆ng RT = 0 ∆H = ∆U = - 13.56kJ mol-1 

Problem 6.9 The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2 (g) and H2O (1) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ∆f H 0 of benzene. Standard enthalpies of formation of CO2 (g) and H O(l) 2 are –393.5 kJ mol–1 and – 285.83 kJ mol–1 respectively.