## 1) Problem Identification **Chemistry — Chemical Equilibrium (Le Chatelier’s principle; effect of adding inert gas).** Reaction: $$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$$ Key idea: **Adding an inert gas at constant $T$ and $P$ changes volume**, hence changes **partial pressures/concentrations**, which can shift equilibrium if $\Delta n_g \neq 0$. --- ## 2) Analysis ### Given 1. A gaseous equilibrium mixture of $$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$$ 2. System is at equilibrium at some fixed $T$. 3. **Xenon ($\mathrm{Xe}$)** is added, and **$T$ and total pressure $P$ are kept constant**. 4. Xenon is **inert** (does not react). ### To find What happens to the equilibrium concentrations (or equivalently partial pressures) of species like $[\mathrm{PCl_3}]$, $[\mathrm{Cl_2}]$, etc., after adding $\mathrm{Xe}$ at constant $T$ and $P$. ### Principles / formulas used 1. Ideal gas law: $$PV = nRT$$ 2. Equilibrium constant in terms of partial pressures: $$K_p = \frac{P_{\mathrm{PCl_3}} \, P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}$$ 3. Reaction mole change: $$\Delta n_g = (1+1) - 1 = +1$$ --- ## 3) Solution Steps ### Step 1: Understand what “adding inert gas at constant $T$ and $P$” does At constant $T$ and total pressure $P$, adding $\mathrm{Xe}$ increases total moles $n_{\text{total}}$. From $$V = \frac{nRT}{P}$$ since $T$ and $P$ are constant and $n$ increases, **$V$ increases**. So the mixture **expands** (gets diluted). --- ### Step 2: See how partial pressures of reacting gases change immediately after expansion For any gas $i$, $$P_i = y_i P$$ where $y_i$ is mole fraction. When $\mathrm{Xe}$ is added, total moles increase, so mole fractions $y_{\mathrm{PCl_5}}$, $y_{\mathrm{PCl_3}}$, $y_{\mathrm{Cl_2}}$ **all decrease** (because now $\mathrm{Xe}$ takes some fraction). Since $P$ is fixed, each reacting gas partial pressure **decreases**. Equivalently, because volume increases, partial pressures of the reacting gases drop by the same dilution factor. Let all reacting-gas partial pressures get multiplied by a factor $\alpha$, where $$0 < \alpha < 1$$ So immediately after adding $\mathrm{Xe}$ (before the reaction shifts): $$P'_{\mathrm{PCl_5}} = \alpha P_{\mathrm{PCl_5}}, \quad P'_{\mathrm{PCl_3}} = \alpha P_{\mathrm{PCl_3}}, \quad P'_{\mathrm{Cl_2}} = \alpha P_{\mathrm{Cl_2}}$$ --- ### Step 3: Compare $Q_p$ right after adding $\mathrm{Xe}$ with $K_p$ Initially at equilibrium: $$Q_p = K_p = \frac{P_{\mathrm{PCl_3}} P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}$$ Immediately after dilution: $$Q_p' = \frac{P'_{\mathrm{PCl_3}} P'_{\mathrm{Cl_2}}}{P'_{\mathrm{PCl_5}}}$$ Substitute the $\alpha$ relations: $$Q_p' = \frac{(\alpha P_{\mathrm{PCl_3}})(\alpha P_{\mathrm{Cl_2}})}{\alpha P_{\mathrm{PCl_5}}}$$ $$Q_p' = \alpha^{(2-1)} \frac{P_{\mathrm{PCl_3}} P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}$$ $$Q_p' = \alpha K_p$$ Since $0 < \alpha < 1$, $$Q_p' < K_p$$ --- ### Step 4: Predict the direction of shift If $$Q_p' < K_p$$ the system must **increase $Q_p$** to return to equilibrium. From $$Q_p = \frac{P_{\mathrm{PCl_3}} P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}$$ to increase $Q_p$, the reaction proceeds **forward** (producing more $\mathrm{PCl_3}$ and $\mathrm{Cl_2}$). This matches Le Chatelier’s principle: at constant $P$, adding inert gas increases volume, and equilibrium shifts toward the side with **more moles of gas**. Here products have $2$ moles vs reactant $1$ mole, so it shifts **to the right**. --- ### Step 5: Conclude concentration changes asked in the key Because the equilibrium shifts forward: - $[\mathrm{PCl_3}]$ **increases** - $[\mathrm{Cl_2}]$ **increases** - $[\mathrm{PCl_5}]$ **decreases** So the statement “$[\mathrm{PCl_3}]$ and $[\mathrm{Cl_2}]$ increase” is correct (Answer key: **C**). --- ## 4) Final Answer and Summary **Final Answer:** On adding $\mathrm{Xe}$ at constant $T$ and $P$, the equilibrium shifts forward, so **$[\mathrm{PCl_3}]$ and $[\mathrm{Cl_2}]$ increase** (and $[\mathrm{PCl_5}]$ decreases). **Summary:** Adding inert gas at constant $T$ and $P$ increases volume, lowering partial pressures and making $Q_p' < K_p$. Since $\Delta n_g = +1$, the system shifts to the product side to restore equilibrium, increasing $\mathrm{PCl_3}$ and $\mathrm{Cl_2}$. **Follow-up question:** If instead $\mathrm{Xe}$ were added at constant $T$ and **constant volume**, would the equilibrium shift? Why?## 1) Problem Identification **Chemistry — Chemical Equilibrium (Le Chatelier’s principle; effect of adding inert gas).** Reaction: $$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$$ Key idea: **Adding an inert gas at constant $T$ and $P$ changes volume**, hence changes **partial pressures/concentrations**, which can shift equilibrium if $\Delta n_g \neq 0$. --- ## 2) Analysis ### Given 1. A gaseous equilibrium mixture of $$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$$ 2. System is at equilibrium at some fixed $T$. 3. **Xenon ($\mathrm{Xe}$)** is added, and **$T$ and total pressure $P$ are kept constant**. 4. Xenon is **inert** (does not react). ### To find What happens to the equilibrium concentrations (or equivalently partial pressures) of species like $[\mathrm{PCl_3}]$, $[\mathrm{Cl_2}]$, etc., after adding $\mathrm{Xe}$ at constant $T$ and $P$. ### Principles / formulas used 1. Ideal gas law: $$PV = nRT$$ 2. Equilibrium constant in terms of partial pressures: $$K_p = \frac{P_{\mathrm{PCl_3}} \, P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}$$ 3. Reaction mole change: $$\Delta n_g = (1+1) - 1 = +1$$ --- ## 3) Solution Steps ### Step 1: Understand what “adding inert gas at constant $T$ and $P$” does At constant $T$ and total pressure $P$, adding $\mathrm{Xe}$ increases total moles $n_{\text{total}}$. From $$V = \frac{nRT}{P}$$ since $T$ and $P$ are constant and $n$ increases, **$V$ increases**. So the mixture **expands** (gets diluted). --- ### Step 2: See how partial pressures of reacting gases change immediately after expansion For any gas $i$, $$P_i = y_i P$$ where $y_i$ is mole fraction. When $\mathrm{Xe}$ is added, total moles increase, so mole fractions $y_{\mathrm{PCl_5}}$, $y_{\mathrm{PCl_3}}$, $y_{\mathrm{Cl_2}}$ **all decrease** (because now $\mathrm{Xe}$ takes some fraction). Since $P$ is fixed, each reacting gas partial pressure **decreases**. Equivalently, because volume increases, partial pressures of the reacting gases drop by the same dilution factor. Let all reacting-gas partial pressures get multiplied by a factor $\alpha$, where $$0 < \alpha < 1$$ So immediately after adding $\mathrm{Xe}$ (before the reaction shifts): $$P'_{\mathrm{PCl_5}} = \alpha P_{\mathrm{PCl_5}}, \quad P'_{\mathrm{PCl_3}} = \alpha P_{\mathrm{PCl_3}}, \quad P'_{\mathrm{Cl_2}} = \alpha P_{\mathrm{Cl_2}}$$ --- ### Step 3: Compare $Q_p$ right after adding $\mathrm{Xe}$ with $K_p$ Initially at equilibrium: $$Q_p = K_p = \frac{P_{\mathrm{PCl_3}} P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}$$ Immediately after dilution: $$Q_p' = \frac{P'_{\mathrm{PCl_3}} P'_{\mathrm{Cl_2}}}{P'_{\mathrm{PCl_5}}}$$ Substitute the $\alpha$ relations: $$Q_p' = \frac{(\alpha P_{\mathrm{PCl_3}})(\alpha P_{\mathrm{Cl_2}})}{\alpha P_{\mathrm{PCl_5}}}$$ $$Q_p' = \alpha^{(2-1)} \frac{P_{\mathrm{PCl_3}} P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}$$ $$Q_p' = \alpha K_p$$ Since $0 < \alpha < 1$, $$Q_p' < K_p$$ --- ### Step 4: Predict the direction of shift If $$Q_p' < K_p$$ the system must **increase $Q_p$** to return to equilibrium. From $$Q_p = \frac{P_{\mathrm{PCl_3}} P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}$$ to increase $Q_p$, the reaction proceeds **forward** (producing more $\mathrm{PCl_3}$ and $\mathrm{Cl_2}$). This matches Le Chatelier’s principle: at constant $P$, adding inert gas increases volume, and equilibrium shifts toward the side with **more moles of gas**. Here products have $2$ moles vs reactant $1$ mole, so it shifts **to the right**. --- ### Step 5: Conclude concentration changes asked in the key Because the equilibrium shifts forward: - $[\mathrm{PCl_3}]$ **increases** - $[\mathrm{Cl_2}]$ **increases** - $[\mathrm{PCl_5}]$ **decreases** So the statement “$[\mathrm{PCl_3}]$ and $[\mathrm{Cl_2}]$ increase” is correct (Answer key: **C**). --- ## 4) Final Answer and Summary **Final Answer:** On adding $\mathrm{Xe}$ at constant $T$ and $P$, the equilibrium shifts forward, so **$[\mathrm{PCl_3}]$ and $[\mathrm{Cl_2}]$ increase** (and $[\mathrm{PCl_5}]$ decreases). **Summary:** Adding inert gas at constant $T$ and $P$ increases volume, lowering partial pressures and making $Q_p' < K_p$. Since $\Delta n_g = +1$, the system shifts to the product side to restore equilibrium, increasing $\mathrm{PCl_3}$ and $\mathrm{Cl_2}$. **Follow-up question:** If instead $\mathrm{Xe}$ were added at constant $T$ and **constant volume**, would the equilibrium shift? Why?

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JEE Mains JEE Advance BITSAT
JEE Mains | PYQ