In an evacuated rigid vessel of volume V litre, one mole of solid ammonium carbonate, NH2COONH4, is — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the balanced decomposition reaction \[\ce{NH2COONH4(s) <=> 2NH3(g) + CO2(g)}\] Step 2 - Set up the ICE table Let \(\alpha\) = moles of solid that dissociate (starting from 1 mol): \[\begin{array}{lccc} & \ce{NH2COONH4(s)} & \ce{2NH3(g)} & \ce{CO2(g)} \ \hline \text{Initial} & 1 & 0 & 0 \ \text{Change} & -\alpha & +2\alpha & +\alpha \ \text{Equilibrium} & 1-\alpha & 2\alpha & \alpha \ \hline \end{array}\] Step 3 - Find total gas moles Solid does not contribute to gas pressure. Total gaseous moles: \[n_{\text{gas}} = 2\alpha + \alpha = 3\alpha\] Step 4 - Apply ideal gas law to find \(\alpha\) \[PV = n_{\text{gas}}RT = 3\alpha RT\] \[\alpha = \frac{PV}{3RT}\] Step 5 - Calculate percentage dissociation \[\% \text{ dissociation} = \alpha \times 100 = \frac{100PV}{3RT}\%\] \[\boxed{\frac{100PV}{3RT}\%}\] Step 6 - Evaluate all options - **Option (A) 100PV/(3RT) %**: Correct. Each mole of solid produces 3 moles of gas, giving \(\alpha = PV/(3RT)\) and percentage = 100PV/(3RT) %. - **Option (B) PV/(3RT) %**: Incorrect. This is the fractional dissociation \(\alpha\), not the percentage. - **Option (C) 100PV/(RT) %**: Incorrect. Assumes only 1 mole of gas per mole dissociated. - **Option (D) 300PV/(RT) %**: Incorrect. Stoichiometric factor of 3 placed in numerator instead of denominator.