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Question: The reaction of cyanamide, NH₂CN(s) with oxygen was run in a bomb calorimeter and ΔU was found to be...

The reaction of cyanamide, NH₂CN(s) with oxygen was run in a bomb calorimeter and ΔU was found to be –742.24 kJ mol⁻¹. The magnitude of ΔH₂₉₈ for the reaction

NH₂CN(s) + 32\frac{3}{2}O₂(g) → N₂(g) + O₂(g) + H₂O(l) is ____ kJ.

A

741.00

B

742.24

C

743.48

D

740.00

Answer

741.00

Explanation

Solution

The relationship between enthalpy change (ΔH\Delta H) and internal energy change (ΔU\Delta U) is given by ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT. For the given reaction, Δng=(moles of gaseous products)−(moles of gaseous reactants)=(1+1)−32=2−1.5=0.5\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = (1 + 1) - \frac{3}{2} = 2 - 1.5 = 0.5 mol. Given ΔU=−742.24\Delta U = -742.24 kJ mol⁻¹, T=298T = 298 K, and R=8.314×10−3R = 8.314 \times 10^{-3} kJ mol⁻¹ K⁻¹. ΔH=−742.24 kJ mol−1+(0.5 mol)(8.314×10−3 kJ mol−1 K−1)(298 K)\Delta H = -742.24 \text{ kJ mol}^{-1} + (0.5 \text{ mol})(8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{ K}^{-1})(298 \text{ K}) ΔH=−742.24 kJ mol−1+1.238786 kJ mol−1\Delta H = -742.24 \text{ kJ mol}^{-1} + 1.238786 \text{ kJ mol}^{-1} ΔH≈−741.001214 kJ mol−1\Delta H \approx -741.001214 \text{ kJ mol}^{-1}. The magnitude of ΔH\Delta H is approximately 741.00 kJ.