Question

temperature.
a(g) + b(g) ⇌ 2c(g) kp = 1.56
d(g) + 3a(g) ⇌ 2e(g) kp = 22.9
what would kp be for the following reaction at the same temperature?
2e(g) + 3b(g) ⇌ 6c(g) + d(g)
a. 4.98 b. 0.180 c. 0.131 d. 0.166 e. 0.0405

Explanation:

Step1: Manipulate the given reactions

Given reaction 1: \(A(g)+B(g)
ightleftharpoons 2C(g)\), \(K_{p1} = 1.56\)
Multiply reaction 1 by 3: \(3A(g)+3B(g)
ightleftharpoons 6C(g)\), \(K_{p1}^{'}=K_{p1}^{3}=(1.56)^{3}\)

Given reaction 2: \(D(g)+3A(g)
ightleftharpoons 2E(g)\), \(K_{p2}=22.9\)
Reverse reaction 2: \(2E(g)
ightleftharpoons D(g)+3A(g)\), \(K_{p2}^{'}=\frac{1}{K_{p2}}=\frac{1}{22.9}\)

Step2: Combine the manipulated reactions

Add the two manipulated reactions:
\((3A(g)+3B(g)
ightleftharpoons 6C(g))+(2E(g)
ightleftharpoons D(g)+3A(g))\) gives \(2E(g)+3B(g)
ightleftharpoons 6C(g)+D(g)\)

The equilibrium constant \(K_p\) for the combined reaction is \(K_p = K_{p1}^{'}\times K_{p2}^{'}\)

Step3: Calculate \(K_p\)

\(K_{p1}^{'}=(1.56)^{3}=1.56\times1.56\times1.56 = 3.796416\)
\(K_p=\frac{(1.56)^{3}}{22.9}=\frac{3.796416}{22.9}\approx0.166\)

Answer:

D. \(0.166\)