500\ mL of nitrogen gas and 750\ mL hydrogen gas were allowed to...

The given reaction is $\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)}$.

Initial volume of $\text{N}_2 = 500\ \text{mL}$
Initial volume of $\text{H}_2 = 750\ \text{mL}$
Total initial volume $V_i = 500 + 750 = 1250\ \text{mL}$

From the stoichiometry, $1$ volume of $\text{N}_2$ reacts with $3$ volumes of $\text{H}_2$ to produce $2$ volumes of $\text{NH}_3$.

Here, $\text{H}_2$ is the limiting reagent. $750\ \text{mL}$ of $\text{H}_2$ will react with $250\ \text{mL}$ of $\text{N}_2$ to form $500\ \text{mL}$ of $\text{NH}_3$.

Volume of $\text{N}_2$ remaining $= 500 - 250 = 250\ \text{mL}$
Total final volume $V_f = 250\ (\text{N}_2) + 500\ (\text{NH}_3) = 750\ \text{mL}$

Change in volume $\Delta V = V_f - V_i = 750 - 1250 = -500\ \text{mL} = -0.5\ \text{L}$

Work done $W = -P_{ext} \Delta V$

$W = -1\ \text{bar} \times (-0.5\ \text{L}) = +0.5\ \text{L}\ \text{bar}$

Since $1\ \text{L}\ \text{bar} = 100\ \text{J}$,

$W = +0.5 \times 100\ \text{J} = +50\ \text{J}$

Answer: $+50\ \text{J}$