For the following reaction at $250^{o} C$ , the value of $K_{c}$ is $26$ , then the value of $K_{p}$ at the same temperature will be:
$P {C l}_{3} (g) + {C l}_{2} (g) ⇌ P {C l}_{5} (g)$

0.57

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0.61

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0.83

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0.91

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Solution

The correct option is B $0.61$
$P {C l}_{3} (g) + {C l}_{2} (g) ⇌ P {C l}_{5} (g)$
$Δ n = 1 - (1 + 1) = - 1$
The value of $K_{c}$ is $26$ at $250^{o} C$ , then the value of $K_{p}$ at the same temperature will be
$K_{p} = K_{c} (R T)^{Δ n}$
$K_{p} = 26 \times (0.08206 L a t m / m o l / K \times (250 + 273.15) K)^{- 1}$
$K_{p} = 0.61$

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For the following reaction at 250oC, the value of Kc is 26, then the value of Kp at the same temperature will be:PCl3(g)+Cl2(g)⇌PCl5(g)

The correct option is B 0.61PCl3(g)+Cl2(g)⇌PCl5(g)Δn=1−(1+1)=−1The value of Kc is 26 at 250oC, then the value of Kp at the same temperature will beKp=Kc(RT)ΔnKp=26×(0.08206Latm/mol/K×(250+273.15)K)−1Kp=0.61

For the following reaction at $250^{o} C$ , the value of $K_{c}$ is $26$ , then the value of $K_{p}$ at the same temperature will be:
$P {C l}_{3} (g) + {C l}_{2} (g) ⇌ P {C l}_{5} (g)$