Question

The equilibrium constant for the equilibrium \(PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)\) at a particular temperature is \(2\times10^{-2}\ \text{mol dm}^{-3}\). 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

• A. 2.0
• B. 2.2
• C. 1.8
• D. 0.2
• E. 0.1

Hint:

For decomposition equilibrium, first find equilibrium concentrations, then add the decomposed amount to get the initial amount.

Answer 1

The Correct Option is B

Concept: For: \[ PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g) \] \[ K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]} \]

Step 1: Use given chlorine concentration.
At equilibrium: \[ [Cl_2]=0.20 \] Since \(PCl_3\) and \(Cl_2\) are formed in equal amount: \[ [PCl_3]=0.20 \]

Step 2: Find equilibrium concentration of \(PCl_5\).
\[ 2\times10^{-2}=\frac{0.20\times0.20}{[PCl_5]} \] \[ [PCl_5]=\frac{0.04}{0.02}=2.0 \]

Step 3: Find initial moles of \(PCl_5\).
Initial \(PCl_5\) must include the amount left plus the amount decomposed: \[ 2.0+0.20=2.2 \]