Answer:
Step 1
H2(g) → 2 H(g)
Cl2(g) → 2 Cl(g)
Step 2
2 H (g) + 2 Cl (g) → 2 HCl
ΔH1 = +(436 kJ + 243 kJ) = +679 kJ
ΔH2 = -2(431 kJ) = -862 kJ
By applying Hess's Law, ΔH = ΔH1 + ΔH2
ΔH = +679 kJ - 862 kJ
ΔH = -183 kJ
Explanation:
Estimate the change in enthalpy, ΔH, 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:
Step 1 The reactant molecules, H2 and Cl2, break down into their atoms.
H2(g) → 2 H(g)
Cl2(g) → 2 Cl(g)
Step 2 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:
ΔH1 = +(436 kJ + 243 kJ) = +679 kJ
ΔH2 = -2(431 kJ) = -862 kJ
By applying Hess's Law, ΔH = ΔH1 + ΔH2
ΔH = +679 kJ - 862 kJ
ΔH = -183 kJ
