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Exercises · 5.8

Q.The reaction of cyanamide, NH2CN(s)NH_2CN(s), with dioxygen was carried out in a bomb calorimeter, and ΔU\Delta U was found to be –742.7 kJ mol−1^{-1} at 298 K. Calculate the enthalpy change for the reaction at 298 K. NH2CN(g)+32O2(g)→N2(g)+CO2(g)+H2O(l)NH_2CN(g) + \tfrac{3}{2} O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)

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The bomb calorimeter gives ΔU=−742.7 kJ mol−1\Delta U = -742.7\ \text{kJ mol}^{-1}; convert with ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT. For this reaction Δng=+12\Delta n_g = +\tfrac{1}{2}, giving ΔH=−741.5 kJ mol−1\Delta H = -741.5\ \text{kJ mol}^{-1}.

Why ΔU≠ΔH\Delta U \ne \Delta H

A bomb calorimeter is rigid, so it measures heat at constant volume, which equals ΔU\Delta U. Enthalpy (constant pressure) is related by

ΔH=ΔU+ΔngRT,\Delta H = \Delta U + \Delta n_g RT,

where Δng\Delta n_g is the change in moles of gas (solids and liquids are neglected).

The reaction

NH2CN(s)+32 O2(g)→N2(g)+CO2(g)+H2O(l),ΔU=−742.7 kJ mol−1.\text{NH}_2\text{CN}(s) + \tfrac{3}{2}\,\text{O}_2(g) \rightarrow \text{N}_2(g) + \text{CO}_2(g) + \text{H}_2\text{O}(l),\qquad \Delta U = -742.7\ \text{kJ mol}^{-1}.

Cyanamide is the solid burnt in the bomb, and water is liquid — neither counts toward Δng\Delta n_g.

Change in moles of gas

  • Gaseous products: N2+CO2=1+1=2 mol\text{N}_2 + \text{CO}_2 = 1 + 1 = 2\ \text{mol}
  • Gaseous reactants: 32 O2=1.5 mol\tfrac{3}{2}\,\text{O}_2 = 1.5\ \text{mol}

Δng=2−32=+12 mol.\Delta n_g = 2 - \tfrac{3}{2} = +\tfrac{1}{2}\ \text{mol}.

Convert to ΔH\Delta H

With R=8.314 J mol−1K−1R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} and T=298 KT = 298\ \text{K}: …

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