Question

PCl\(_5\)(g), PCl\(_3\)(g) and Cl\(_2\)(g) are at equilibrium at 500 K. The equilibrium concentrations of PCl\(_3\), Cl\(_2\) and PCl\(_5\) are respectively 4.0 M, 4.0 M and 2.0 M. Calculate \( K_c \) for the reaction: \[ {PCl}_5(g) \rightleftharpoons {PCl}_3(g) + {Cl}_2(g) \]

• A. 2 mol dm\(^3\)
• B. 4 mol dm\(^3\)
• C. 6 mol dm\(^3\)
• D. 8 mol dm\(^3\)
• E. 10 mol dm\(^3\)

Hint:

Remember, for equilibrium expressions, concentrations of products are multiplied and divided by the concentrations of reactants raised to their respective stoichiometric coefficients.

Answer 1

The Correct Option is D

The expression for \( K_c \) is: \[ K_c = \frac{[{PCl}_3][{Cl}_2]}{[{PCl}_5]} \] Substitute the given equilibrium concentrations: \[ K_c = \frac{(4.0)(4.0)}{2.0} = \frac{16}{2} = 8 \, {mol dm}^{-3} \] Thus, the correct answer is (D).