Ammonia at a pressure of 10 atm and at a pressure of 20 atm are introduced into an evacuated chamber — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Equilibrium and Identify Initial Conditions Heterogeneous decomposition of solid ammonium carbamate: \[\ce{NH2COONH4(s) <=> 2NH3(g) + CO2(g)}, \quad K_p = 2020\text{ atm}^3\] Initial conditions in the evacuated chamber: * $p_{\ce{NH3},0} = 10\text{ atm}$, $p_{\ce{CO2},0} = 20\text{ atm}$ * **No solid \ce{NH2COONH4} present** (only gases were introduced) Step 2 - Calculate the Reaction Quotient (\(Q_p\)) \[Q_p = (p_{\ce{NH3},0})^2 \times p_{\ce{CO2},0} = (10)^2 \times 20 = 2000\text{ atm}^3\] Step 3 - Compare \(Q_p\) with \(K_p\) and Analyze Reaction Paths $Q_p = 2000 < K_p = 2020$ Analysis of two possible reaction directions: **Forward reaction (solid decomposing → more gas):** Since $Q_p < K_p$, thermodynamically the forward direction is favored. But the forward reaction requires solid \ce{NH2COONH4}. Since **no solid is present**, the forward reaction cannot occur. **Backward reaction (gases → solid):** For the reverse reaction to occur spontaneously, we need $Q_p > K_p$. Since $Q_p = 2000 < K_p = 2020$, the backward reaction is thermodynamically non-spontaneous and cannot occur. Step 4 - Conclude the Total Pressure Since neither direction is accessible, the system remains unchanged: \[P_{\text{total}} = p_{\ce{NH3}} + p_{\ce{CO2}} = 10 + 20 = \boxed{30\text{ atm}}\] Step 5 - Explain Each Option * **(A) less than 30 atm**: Incorrect. Total pressure decreases only if the reverse reaction (consuming gas) occurs, but $Q_p < K_p$ prevents it. * **(B) more than 30 atm**: Incorrect. Total pressure increases only if the forward reaction (producing gas) occurs, but no solid reactant is present. * **(C) equal to 30 atm**: Correct. Neither reaction can proceed; total pressure stays at 30 atm. * **(D) unpredictable**: Incorrect. The outcome is clearly predictable using thermodynamics.