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The value of Kp for the reaction 2 A(g) + B(g) + 3 C(g) → 2 D(g) + E(g) is 18530 at a particular temperature. What would be the value of Kp for the reaction 2 D(g) + E(g) → 2 A(g) + B(g) + 3 C(g)?

1 Answer

5 votes

Answer:


Kp2=5.3967x10^(-5)

Step-by-step explanation:

Hello there!

In this case, since the Kp for the first reaction is given, according to the following equilibrium expression:


Kp=(p_D^2p_E)/(p_A^2p_Bp_C^3) =18530

However, since the second reaction stands for the reverse of the initial one, the equilibrium expression would be:


Kp_2=Kp=(p_A^2p_Bp_C^3)/(p_D^2p_E)

And therefore its Kp the inverse of the aforementioned one:


Kp2=1/18530\\\\Kp2=5.3967x10^(-5)

Best regards!

User Sanpaco
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