Question

\(500\ \text{mL}\) of nitrogen gas and \(750\ \text{mL}\) hydrogen gas were allowed to react at \(1\ \text{bar}\) pressure according to reaction,
\(\text{N}_{2(g)}+3\text{H}_{2(g)}\rightarrow 2\text{NH}_{3(g)}\)
What is PV type of work done during reaction?

• A. \(+50\ \text{J}\)
• B. \(-50\ \text{J}\)
• C. \(+100\ \text{J}\)
• D. \(-100\ \text{J}\)

Hint:

Find the change in gas volume from the reaction ratio, then use w = -P times change in volume.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
At constant pressure the PV work is \(w=-P\Delta V\). Gases at the same temperature and pressure react in the ratio of their mole coefficients.

Step 2: Find the limiting reactant
\(\text{N}_{2}:\text{H}_{2}=1:3\). 750 mL of \(\text{H}_{2}\) needs 250 mL of \(\text{N}_{2}\), so \(\text{H}_{2}\) is limiting and gives 500 mL of \(\text{NH}_{3}\).

Step 3: Volume change
Initial volume \(=500+750=1250\ \text{mL}\).
Final volume \(=250\ (\text{N}_{2}\text{ left})+500\ (\text{NH}_{3})=750\ \text{mL}\).
\(\Delta V=-500\ \text{mL}=-0.5\ \text{L}\)

Step 4: Work done
\(w=-P\Delta V=-(1\ \text{bar})(-0.5\ \text{L})=+0.5\ \text{bar L}\)
Since \(1\ \text{bar L}=100\ \text{J}\), \(w=+50\ \text{J}\).

Final Answer:
The PV work done on the system is plus 50 J. \[ \boxed{\text{(A) }+50\ \text{J}} \]