Question

Chemical energy is released or absorbed from reactions in various forms. The most easily measurable form of energy comes in the form of heat, or enthalpy. The enthalpy of a reaction can be calculated from the heats of formation of the substances involved in the reaction:

ΔH∘rxn=Σ npΔH∘f(products)−Σ nrΔH∘f(reactants)

where n represents the stoichiometric coefficients.

A:

Calculate the standard enthalpy change for the reaction

2A+B⇌2C+2D

Use the following data:

Substance	ΔH∘f (kJ/mol)
A	-253
B	-411
C	177
D	-507

B: For the reaction given in Part A, how much heat is absorbed when 3.00 mol of A reacts?

Answer 1

A)

Given:

Hof(A) = -253.0 KJ/mol

Hof(B) = -411.0 KJ/mol

Hof(C) = 177.0 KJ/mol

Hof(D) = -507.0 KJ/mol

Balanced chemical equation is:

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

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

ΔHo rxn = 2*(177.0) + 2*(-507.0) - 2*(-253.0) - 1*(-411.0)

ΔHo rxn = 257 KJ

Answer: 257 KJ

B)

ΔHo rxn = 257 KJ is when 2 mol of A reacts as per equation.

For 3.00 mol of A,

ΔHo rxn = 257 * 3 / 2 = 385.5 KJ

Answer: 385.5 KJ