Solid ammonium carbamate dissociates: NH2COONH4(s) ⇌ 2(g) + (g). At equilibrium, is added such that — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Find Kp at original equilibrium The dissociation reaction: \[\ce{NH2COONH4(s) <=> 2NH3(g) + CO2(g)}\] Let \(p\) = partial pressure of \ce{CO2} at original equilibrium. By stoichiometry: \(p_{\ce{NH3}} = 2p\), original total pressure = \(3p\). \[K_p = p_{\ce{NH3}}^2 \cdot p_{\ce{CO2}} = (2p)^2 \cdot p = 4p^3\] Step 2 - Find new equilibrium after adding NH3 The problem states that at new equilibrium, \(p'_{\ce{NH3}} = 3p\) (equals original total pressure). Since \(K_p\) is constant (temperature unchanged): \[K_p = (p'_{\ce{NH3}})^2 \cdot p'_{\ce{CO2}} = (3p)^2 \cdot p'_{\ce{CO2}} = 9p^2 \cdot p'_{\ce{CO2}} = 4p^3\] \[p'_{\ce{CO2}} = \frac{4p^3}{9p^2} = \frac{4p}{9}\] Step 3 - Calculate ratio of pressures New total pressure: \[P'_\text{total} = p'_{\ce{NH3}} + p'_{\ce{CO2}} = 3p + \frac{4p}{9} = \frac{27p + 4p}{9} = \frac{31p}{9}\] Original total pressure: \[P_\text{orig} = 3p = \frac{27p}{9}\] \[\frac{P'_\text{total}}{P_\text{orig}} = \frac{31p/9}{27p/9} = \boxed{\frac{31}{27}}\] The ratio of new total pressure to original total pressure is \(\mathbf{31 : 27}\) (option C). Step 4 - Evaluate all options - **Option (A) 1:1**: Incorrect. Adding NH3 shifts equilibrium backward, reducing CO2, so total pressure changes. - **Option (B) 27:31**: Incorrect. This is the inverted ratio. - **Option (C) 31:27**: Correct. As derived, new total = 31p/9 and original = 27p/9. - **Option (D) 3:4**: Incorrect. Does not follow from the Kp constraint.