Question

Part A

Calculate the standard enthalpy change for the reaction

2A+B⇌2C+2D

where the heats of formation are given in the following table:

Substance	ΔH∘f (kJ/mol)
A	-227
B	-399
C	213
D	-503

Express your answer in kilojoules.

Answer= 273kJ

Part B:

For the reaction given in Part A, how much heat is absorbed when 3.70 mol of A reacts?

Express your answer numerically in kilojoules.

Part C:

For the reaction given in Part A, ΔS∘rxn is 25.0 J/K . What is the standard Gibbs free energy of the reaction, ΔG∘rxn?

Answer 1

A)

we have:

Hof(A) = -227.0 KJ/mol

Hof(B) = -399.0 KJ/mol

Hof(C) = 213.0 KJ/mol

Hof(D) = -503.0 KJ/mol

we have the Balanced chemical equation as:

2 A + B ---> 2 C + 2 D

deltaHo rxn = 2*Hof(C) + 2*Hof(D) - 2*Hof( A) - 1*Hof(B)

deltaHo rxn = 2*(213.0) + 2*(-503.0) - 2*(-227.0) - 1*(-399.0)

deltaHo rxn = 273 KJ

Answer: 273 KJ

B)

From above reaction,

when 2 mol of A reacts, deltaHo rxn = 273 KJ

so, for 3.70 mol, deltaHo rxn = 273*3.70/2 = 505 KJ

Answer: 505 KJ

C)

deltaHo = 273.0 KJ/mol

deltaSo = 25 J/mol.K

= 0.025 KJ/mol.K

T = 298 K

we have below equation to be used:

deltaGo = deltaHo - T*deltaSo

deltaGo = 273.0 - 298.0 * 0.025

deltaGo = 265.55 KJ/mol

Answer: 265.55 KJ/mol