Question

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)?

Answer 1

`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!