Problem
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
Bond breaking requires energy, so we expect the value for ΔH to be positive for this step.
In the second step of the reaction, two moles of H-Cl bonds are formed. Bond breaking liberates energy, so we expect the ΔH 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:
ΔH2 = -2(431 kJ) = -862 kJ
By applying Hess's Law, ΔH = ΔH1 + ΔH2
ΔH = +679 kJ - 862 kJ
ΔH = -183 kJ
Answer
The enthalpy change for the reaction will be ΔH = -183 kJ.
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