Question:
The equilibrium constant for the equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ at a particular temperature is $2 \times 10^{-2}$ mol L$^{-1}$. The number of moles of $PCl_5$ that must be taken in a one-litre flask at the same temperature to obtain a concentration of $0.20$ mol of chlorine at equilibrium is}
The equilibrium constant for the equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ at a particular temperature is $2 \times 10^{-2}$ mol L$^{-1}$. The number of moles of $PCl_5$ that must be taken in a one-litre flask at the same temperature to obtain a concentration of $0.20$ mol of chlorine at equilibrium is}
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Always set up ICE table carefully. Mistakes usually happen in equilibrium expressions.
Updated On: May 8, 2026
- $2.2$
- $2.0$
- $1.8$
- $0.2$
- $0.1$
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The Correct Option is A
Solution and Explanation
Concept:
For equilibrium:
\[
K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}
\]
Step 1: Assume initial moles. Let initial moles of $PCl_5 = x$ (since volume = 1 L, concentration = moles)
Step 2: Let dissociation occur. \[ PCl_5 \rightarrow PCl_3 + Cl_2 \] At equilibrium: \[ [PCl_5] = x - 0.2 \] \[ [PCl_3] = 0.2 \] \[ [Cl_2] = 0.2 \]
Step 3: Apply equilibrium constant. \[ 2 \times 10^{-2} = \frac{0.2 \times 0.2}{x - 0.2} \]
Step 4: Solve. \[ 2 \times 10^{-2} = \frac{0.04}{x - 0.2} \] \[ x - 0.2 = \frac{0.04}{2 \times 10^{-2}} = 2 \] \[ x = 2 + 0.2 = 2.2 \]
Step 1: Assume initial moles. Let initial moles of $PCl_5 = x$ (since volume = 1 L, concentration = moles)
Step 2: Let dissociation occur. \[ PCl_5 \rightarrow PCl_3 + Cl_2 \] At equilibrium: \[ [PCl_5] = x - 0.2 \] \[ [PCl_3] = 0.2 \] \[ [Cl_2] = 0.2 \]
Step 3: Apply equilibrium constant. \[ 2 \times 10^{-2} = \frac{0.2 \times 0.2}{x - 0.2} \]
Step 4: Solve. \[ 2 \times 10^{-2} = \frac{0.04}{x - 0.2} \] \[ x - 0.2 = \frac{0.04}{2 \times 10^{-2}} = 2 \] \[ x = 2 + 0.2 = 2.2 \]
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