A reaction vessel is charged with 0.50 atm of A and 0.030 atm of B. Once the reaction reaches equilibrium, what is the equilibrium partial pressure of B? Kp for this reaction is 67.2A (g) ⇌ 2 B (g)

1.000 atm

Explanation:

Step 1: Write the reversible reaction

A(g) ⇌ 2 B(g)

Step 2: Calculate the reaction quotient

Q = [B]²/[A]

Q = 0.030²/0.50

Q = 0.0018

since Q < Kp, the reaction will proceed to the right

Step 3: Make an ICE chart

       A(g) ⇌ 2 B(g)

I      0.50      0.030

C       -x          +2x

E    0.50-x    0.030+2x

Step 4: Find the value of "x"

We will use the definition of the pressure equilibrium constant.

Kp = 67.2 = [B]²/[A] = (0.030+2x)²/0.50-x

33.6 - 67.2x = 0.0009 + 0.12x + 4x²

4x² + 67.32x -33.5991 = 0

We solve this quadratic equation and we get x=0.485112  and x=−17.3151 (neglected).

Step 5: Find the partial pressure of B at equilibrium

pB = 0.030+2x = 0.030+2(0.485112) = 1.000 atm

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explanation: