Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

Given:

• Decomposition of 1 mole of NH₄NO₃
• Temperature \(T = 100 \, ^\circ \text{C} = 373 \, \text{K}\)

To find: Work done and to determine whether work is done on the system or by the system.

Formula: \(W = -\Delta n_g RT\)

Calculation:

• Molar mass of NH₄NO₃ = \(2 \times 14 + 3 \times 16 + 4 \times 1 = 80 \, \text{g mol}^{-1}\)
• Moles of NH₄NO₃ (\(n\)) = \(\frac{132 \, \text{g}}{80 \, \text{g mol}^{-1}} = 1.65 \, \text{mol}\)

The given reaction is for 1 mole of NH₄NO₃. For 1.65 moles of NH₄NO₃, the reaction is given as follows:

1.65 NH₄NO_3(s) → 1.65 N₂O_(g) + 3.30 H₂O_(g)

Now,

\(\Delta n_g = (\text{moles of product gases}) - (\text{moles of reactant gases})\)

\(\Delta n_g = (1.65 + 3.30) - 0 = +4.95 \, \text{mol} \, (\text{since NH₄NO₃ is in solid state})\)

Hence,

\(W = -\Delta n_g RT\)

\(= -(+4.95 \, \text{mol}) \times 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1} \times 373 \, \text{K}\)

\(= -15350 \, \text{J} = -15.35 \, \text{kJ}\)

Work is done by the system (since \(W < 0\)).

The work done is \(-15.35 \, \text{kJ}\). The work is done by the system.