For reaction, PCl3(.g.)+Cl2(.g.)leftharpoons PCl5(.g.) , the value of Kc at 250° C is 26. The value of Kp at this temperature will be

Question

For reaction, PCl3(.g.)+Cl2(.g.)leftharpoons PCl5(.g.) , the value of Kc at 250° C is 26. The value of Kp at this temperature will be (A) 0.61 (B) 0.57 (C) 0.83 (D) 0.46

Tardigrade Admin · Accepted Answer

K p = K c ( RT )Δ n Here K c =26, R =0.0821 L atm mol -1 K -1, T =250+273=523 K Δ n =1-(1+1)=-1 So K p =26 ×(0.0821 × 523)-1=0.605 ≃ 0.61

Q. For reaction, $PC l_{3} (g) + C l_{2} (g) ⇌ PC l_{5} (g)$ , the value of $K_{c}$ at $25 0^{\circ} C$ is 26. The value of $K_{p}$ at this temperature will be

0.61

0.57

0.83

0.46

$K_{p} = K_{c} (RT)^{Δ n}$

Here $K_{c} = 26, R = 0.0821 L$ atm $m o l^{- 1} K^{- 1},$

$T = 250 + 273 = 523 K$

$Δ n = 1 - (1 + 1) = - 1$

So $K_{p} = 26 \times (0.0821 \times 523)^{- 1} = 0.605 ≃ 0.61$

Q. For reaction, PCl3​(g)+Cl2​(g)⇌PCl5​(g) , the value of Kc​ at 250∘C is 26. The value of Kp​ at this temperature will be