Question

Calculate the work done in the following reaction
\(\text{NH}_{4}\text{NO}_{3(s)}\rightarrow \text{N}_{2}\text{O}_{(g)}+2\text{H}_{2}\text{O}_{(g)}\) at \(300\) K.
[\(R=8.314\ \text{J K}^{-1}\ \text{mol}^{-1}\)]

• A. \(-7.483\ \text{kJ mol}^{-1}\)
• B. \(-3.741\ \text{kJ mol}^{-1}\)
• C. \(-11.224\ \text{kJ mol}^{-1}\)
• D. \(-14.966\ \text{kJ mol}^{-1}\)

Hint:

Work done equals minus the change in moles of gas times RT.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For reactions involving gases, \(w=-\Delta n_g RT\), where \(\Delta n_g\) is moles of gaseous products minus gaseous reactants.

Step 2: Find the Change in Moles of Gas
Reactant \(\text{NH}_{4}\text{NO}_{3}\) is a solid, so \(n_g=0\).
Products: \(1+2=3\) mol gas, so \(\Delta n_g=3\).

Step 3: Calculation
\(w=-3\times 8.314\times 300=-7482.6\ \text{J}\approx -7.483\ \text{kJ mol}^{-1}\)

Step 4: Compare the options
This matches option (A).

Final Answer:
The work done is about minus 7.483 kJ per mole. \[ \boxed{\text{(A) }-7.483\ \text{kJ mol}^{-1}} \]