Question

The reaction of cyanamide, NH$_2$CN(s), with oxygen was run in a bomb calorimeter and $\Delta U$ was found to be -742.24 kJ mol$^{-1}$. The magnitude of $\Delta H_{298}$ for the reaction NH$_2$CN(s) + 3/2 O$_2$(g) $\rightarrow$ N$_2$(g) + CO$_2$(g) + H$_2$O(l) is _________ kJ.

Hint:

[Image of bomb calorimeter diagram] A bomb calorimeter measures $\Delta U$ because it is a constant-volume process ($W=0$). For reactions where $\Delta n_g$ is positive, $\Delta H$ will be less negative (higher) than $\Delta U$.

Answer 1

Correct Answer: 741

Step 1: Use the relation $\Delta H = \Delta U + \Delta n_g RT$.
Step 2: Calculate $\Delta n_g$ (gaseous products - gaseous reactants): $\Delta n_g = (1 \text{ [N}_2\text{]} + 1 \text{ [CO}_2\text{]}) - 1.5 \text{ [O}_2\text{]} = 2 - 1.5 = +0.5$.
Step 3: $\Delta H = -742.24 + (0.5 \times 8.314 \times 10^{-3} \times 298)$.
Step 4: $\Delta H = -742.24 + 1.238 = -741.002$ kJ.
Step 5: Magnitude = $741$ kJ.