Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 4HCl(g) + O2(g)—>2H2O(g) + 2Cl2(g)

ΔH°r = -114.40 kJ

Explanation:

Let's consider the following balanced equation.

4 HCl(g) + O₂(g) ⇒ 2 H₂O(g) + 2 Cl₂(g)

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

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where

ΔH°r = 2 mol × ΔH°f(H₂O(g)) + 2 mol × ΔH°f(Cl₂(g)) - 4 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(O₂(g))

ΔH°r = 2 mol × (-241.82 kJ/mol) + 2 mol × 0 kJ/mol - 4 mol × (-92.31 kJ/mol) - 1 mol × 0 kJ/mol

ΔH°r = -114.40 kJ

the answer is:  

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