Q. For the reaction, $PCl_{3}(g) + Cl_{2}(g) \rightleftharpoons PCl_{5}(g)$, the value of $K_{c}$ at $250℃$ is 26. The value of $K_{p}$ at this temperature will be
A) 0.41
B) 0.51
C) 0.61
D) 0.71
Explanation
(c)
$K_{p} = K_{c}(RT)^{\mathrm{\Delta}ng}$
Here, $\mathrm{\Delta}n_{g} = - 1$ and $K_{c} = 26$
$R = 0.0812,\ T = 250 + 273 = 523$ K
$K_{p} = 26 \times (0.0812 \times 523)^{- 1} = 0.605 = 0.61$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.