Question

At T(K), \( K_c \) value for the dissociation of \( \text{PCl}_5 \) is 0.04.
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \] The number of moles of \( \text{PCl}_5 \) to be added to a 3.0 L flask to get chlorine concentration of 0.15 mol L\(^{-1}\) is (approximately)?

• A. 1.7
• B. 1.2
• C. 2.1
• D. 1.8

Hint:

Use the equilibrium constant expression and plug in known values to find the required initial concentration or moles.

Answer 1

The Correct Option is C

Step 1: Let equilibrium concentration of Cl\(_2\) = 0.15 mol/L
Then, equilibrium concentration of PCl\(_3\) = 0.15 mol/L, since 1:1 ratio.
Let initial concentration of PCl\(_5\) be \( x \), so equilibrium concentration of PCl\(_5\) is \( x - 0.15 \). \[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.15)^2}{x - 0.15} = 0.04 \] \[ \Rightarrow \frac{0.0225}{x - 0.15} = 0.04 \Rightarrow x - 0.15 = \frac{0.0225}{0.04} = 0.5625 \Rightarrow x = 0.7125 \] Moles of PCl\(_5\) = concentration \( \times \) volume = \( 0.7125 \times 3 = 2.1375 \approx 2.1 \)