Question

One method for producing hydrogen is to react coke (mostly solid carbon) with steam to make syngas, a mixture of CO and hydrogen

C(s) + H₂O(g) ⇌ CO(g) + H₂(g)     K_p = 0.45 at 900 K     ΔH = +131 kJ

What is the partial pressure of CO at equilibrium when 9.59 atm of H₂O(g) is heated with excess C(s) at 900 K?

After equilibrium is reached, CO(g) is added.

In which direction will the reaction shift?

Will the equilibrium constant, K, increase or decrease?

Answer 1

Answer – We are given, reaction –

C(s) + H₂O(g) <-----> CO(g) + H₂(g)     K_p = 0.45 at 900 K

P of H₂O = 9.59 atm , excess of C(s) , so we need to put ICE chart first

     C(s) + H₂O(g) <-----> CO(g) + H₂(g)

I              9.59                   0             0

C                -x                   +x           +x

E             9.59-x               +x           +x

We know the , Kp expression for this one

Kp = P(CO) * P(H₂) / P(H₂O)

0.45 = x * x / (9.59-x)

0.45 (9.59-x) = x²

4.32 -0.45x =x²

So, x²+0.45x -4.32 = 0

Using the quadratic equation

x = 1.865

so, at the equilibrium the P of CO = x = 1.865 atm .

After the equilibrium when we added the CO(g) is added then according to Le Chatelier's Principle there is equilibrium with shift towards opposite side, means when we CO(g) is added then the equilibrium will shift towards the reactant side, means left side.

As the reaction gets reversed due to CO(g) is added at equilibrium there is equilibrium constant also gets inverse. So the new equilibrium is Kp^’ = 1/ 0.45 = 2.22

So the equilibrium constant K gets increase