One method for producing hydrogen is to react coke (mostly solid carbon) with st

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)     Kp = 0.45 at 900 K     H = +131 kJ

What is the partial pressure of CO at equilibrium when 8.39 atm of H2O(g) is heated with excess C(s) at 900 K?

PCO = ? atm

After equilibrium is reached, H2O(g) is removed.

left, right, no shift: In which direction will the reaction shift?

increase, decrease, no change: Will the equilibrium constant, K, increase or decrease?

Explanation / Answer

Answer – We are given, reaction – C(s) + H2O(g) <-----> CO(g) + H2(g) Kp = 0.45 at 900 K

P of H2O = 8.39 atm , excess of C(s) ,

First we need to put ICE chart first

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

I               8.39                       0             0

C                -x                         +x           +x

E             8.39-x                      +x           +x

Kp expression for this one as follow-

Kp = P(CO) * P(H2) / P(H2O)

0.45 = x * x / (8.39-x)

0.45 (8.39-x) = x2

3.776 -0.45x =x2

So, x2 + 0.45x -3.776 = 0

Using the quadratic equation

x = 1.73

so, at the equilibrium the P of CO = 1.73 atm .

After the equilibrium when we H2O(g) is removed then according to Le Chatelier's principle there is equilibrium with shift towards same side, means when we H2O(g) is removed then the equilibrium will shift towards the reactant side, means left side.

Due to H2O(g) is remove at equilibrium there is reaction gets reversed and equilibrium constant also gets inverse. So the new equilibrium is Kp’ = 1/ 0.45 = 2.22

So the equilibrium constant K gets increase

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.