Estimate the change in enthalpy, DH, for the following reaction:

H₂(g) + Cl₂(g) --> 2 HCl(g)

Solution

To work this problem, think of the reaction in terms of simple steps:

1. First, the reactant molecules, H₂ and Cl₂, break down into their atoms

H₂(g) --> 2 H(g)
Cl₂(g) --> 2 Cl(g)

2. Next, these atoms combine to form HCl molecules

2 H(g) + 2 Cl(g) --> 2 HCl(g)

In the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:

DH₁ = +(436 kJ + 243 kJ) = +679 kJ

Bond breaking requires energy, so we expect the value for DH to be positive for this step.

In the second step of the reaction, two moles are H-Cl bonds are formed. Bond breaking liberates energy, so we expect the DH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:

DH₂ = -2(431 kJ) = -862 kJ

By applying Hess's Law, DH = DH₁ + DH₂

DH = +679 kJ - 862 kJ

DH = -183 kJ

Answer

DH = -183 kJ