1. Given data

• For the combustion reaction in a bomb calorimeter: $\Delta U = -742.24\ \text{kJ mol}^{-1}$
• Temperature: $T = 298\ \text{K}$
• Gas constant: $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} = 0.008314\ \text{kJ mol}^{-1}\text{K}^{-1}$

We need the magnitude of $\Delta H_{298}$.

2. Use the relation between $\Delta H$ and $\Delta U$

For ideal gases, $\Delta H = \Delta U + \Delta n_g RT$ where $\Delta n_g$ is the change in moles of gaseous species.

3. Find $\Delta n_g$ from the given reaction

The reaction is written as: $NH_2CN_{(s)} + \frac{3}{2}O_{2(g)} \to N_{2(g)} + O_{2(g)} + H_2O_{(l)}$

Count gaseous moles:

• Reactant gases = $\frac{3}{2}$
• Product gases = $1 + 1 = 2$

So, $\Delta n_g = 2 - \frac{3}{2} = \frac{1}{2}$

4. Calculate $\Delta n_g RT$

$\Delta n_g RT = \frac{1}{2} \times 0.008314 \times 298$

$= \frac{1}{2} \times 2.477572$

$= 1.238786\ \text{kJ mol}^{-1}$

5. Calculate $\Delta H$

$\Delta H = -742.24 + 1.238786$

$\Delta H = -741.001214\ \text{kJ mol}^{-1}$

Thus, the magnitude is $|\Delta H| = 741.001214\ \text{kJ mol}^{-1}$

Rounded to nearest integer: $\boxed{741}$

6. Comparison with stored answer

Stored correct answer = $741$

Our derived answer also gives $741$, so they agree.