Question

The production of hydrogen chloride gas (HCl) at 1.0 atm and 25°C is represented by the following thermochemical equation.

H₂(g) + Cl₂(g) → 2 HCl(g);    

ΔH = −184.6 kJ

Suppose exactly 4 mol H₂(g) reacts with exactly 4 mol Cl₂(g) to form HCl(g) under the same conditions of temperature and pressure.

(a) Calculate the amount of pressure-volume work that is done.
kJ

(b) Calculate the change in internal energy (ΔE) of the system, assuming the reaction proceeds to completion.
kJ

Answer 1

For the reaction,

H2(g) + Cl2(g) → 2 HCl(g); ΔH = −184.6 kJ

The value of ΔH is negative indicates exothermic reaction.

(a) Calculation of work done

When 4 mol H2(g) reacts with 4 mol Cl2(g) to form HCl(g) under the same conditions of temperature and pressure. Then above reaction is represented by,

4H2(g) + 4Cl2(g) → 8 HCl(g); ΔH = 4x(−184.6 )kJ = −738.4 KJ

Since, there is no increase in moles of gas and the products occupy the same volume as the reactants, hence ΔV = 0

We know that work done = PΔV = P x 0 = 0 (no work is done regardless of the number of moles)

(b) Calculation of Internal energy : For the reaction,

4H2(g) + 4Cl2(g) → 8 HCl(g); ΔH = 4x(−184.6 )kJ =  −738.4 KJ

Enthalpy change (ΔH) is related with change in internal energy ( ΔE) as,

ΔH = ΔE + PΔV

-738.4 KJ = ΔE + 0 (PΔV= work done =0) then,

ΔE = -738.4 KJ