For the reaction $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ the value of $K_{c}$ at $250^{o} C$ is 26. The value of $K_{p}$ at this temperature will be :

0.61

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0.57

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0.83

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0.46

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Solution

The correct option is A 0.61
$K_{p} = K_{c} \times (R T)^{Δ n}$
$Δ n =$ no of moles of gaseous produt - number of moles of gaseous rectant
therefore $Δ n = 1 - 2 = - 1$
R = 0.0821, T = 250 + 273.15 = 523.15 and $K_{c}$ = 26
so $K_{p} = K_{c} \times (R T)^{- 1}$
$K_{p} = \frac{26}{0.0821 \times 523.15} = 0.61$

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For the reaction PCl3(g)+Cl2(g)⇌PCl5(g) the value of Kc at 250oC is 26. The value of Kp at this temperature will be :

The correct option is A 0.61Kp=Kc×(RT)ΔnΔn= no of moles of gaseous produt - number of moles of gaseous rectanttherefore Δn=1−2=−1R = 0.0821, T = 250 + 273.15 = 523.15 and Kc = 26soKp=Kc×(RT)−1 Kp=260.0821×523.15=0.61

For the reaction $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ the value of $K_{c}$ at $250^{o} C$ is 26. The value of $K_{p}$ at this temperature will be :