Question

Calculate the value of Kp for the equation.

C(s) + CO2(g) 2CO(g) Kp=?

Given that at a certain temperature:

C(s) + 2H2O(g) CO2(g) + 2H2(g) Kp1= 3.53

H2(g) + CO2(g) H2O(g) + CO(g) Kp2 = 0.699

Kp=?????

Answer 1

We need find the equilibrium constant expressed in terms of partial pressures ie. Kp for above asked question using both equations as:

C_s    + CO_2g   ======= 2CO_g   --------eq1

C_s    + 2H₂O_g   ======= CO_2g   +     2H_2g           Kp¹ = 3.53    ---eq2

H_2g +   CO_2g   =======    H2Og +     CO_g           Kp² = 0.699     ---eq3

In order to do so, we need to find a way to express the first reaction (eq1) in terms of the two reactions for which the equilibrium constant is known or given (from Hess` law)

Notice that happens when we multiply the eq3 by 2

                [H_2g   +   CO_2g     ======   H₂O_g    +   CO_g]×2

So,         2H_2g   +   2CO_2g =======   2H₂O_g   +    2CO_g

The equilibrium constant for this reaction is:

       K′p2 = (H₂O)^₂⋅(CO)² / (H₂)²⋅(CO₂)²

But, since the original reaction had:

Kp2 = (H₂O)⋅(CO) / (H₂)⋅(CO₂)

so we can say:

K′p2 =   [(H₂O)⋅(CO) / (H₂)⋅(CO₂)]² = Kp2²

After adding this reaction to eq3 and comparing eq2, a net reaciton comes as:

C_s   +   2CO_2g ====== CO_2g     +    2CO_g

This is equivalent to our eq1

The equilibrium constant will be equal to

Kp = Kp1× K′p2

Kp = Kp1 × Kp2²

Kp = 3.53 × 0.699² = 3.53 x 0.488601

Kp = 1.724