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

Answer:

$Kp2=5.3967x10^{-5}$

Explanation:

Hello there!

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

$Kp=\frac{p_D^2p_E}{p_A^2p_Bp_C^3} =18530$

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

$Kp_2=Kp=\frac{p_A^2p_Bp_C^3}{p_D^2p_E}$

And therefore its Kp the inverse of the aforementioned one:

$Kp2=1/18530\\\\Kp2=5.3967x10^{-5}$

Best regards!