Question

Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H2(g) with 1 mole of Cl2(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

Answer 1

Answer:

ΔH°r = -184.6 kJ

Explanation:

Let's consider the following balanced equation.

H₂(g) + Cl₂(g) ⇄ 2 HCl(g)

We can calculate the standard enthalpy of the reaction (ΔH°r) using the following expression:

ΔH°r = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(p) are the standard enthalpies of formation of reactants and products

By definition, the standard enthalpy of formation of a simple substance in its most stable state is zero. Then,

ΔH°r = 2 mol × ΔH°f(HCl(g)) - [1 mol × ΔH°f(H₂(g)) + 1 mol × ΔH°f(Cl₂(g))]

ΔH°r = 2 mol × (-92.3 kJ/mol) - [1 mol × 0 + 1 mol × 0]

ΔH°r = -184.6 kJ