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.

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

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