Question

At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 ´ 10-7. 2 H2S(g) f 2 H2(g) + S2(g) A reaction vessel at 800 K initially contains 3.50 atm of H2S. If the reaction is allowed to equilibrate, what is the equilibrium pressure of H2?

Answer 1

The given equilibrium reaction is

Initially, we have 3.50 atm of reactant H₂S.

Hence, the reaction will proceed in forward direction.

Now, we can create an ICE table to calculate the equilibrium partial pressures.

Initial, atm	3.50	0	0
Change, atm	-2x	+2x	+x
Equilibrium , atm	3.50-2x	2x	x

Note that for -2x change in partial pressure of H₂S, we get a change of +2x in H₂ and +x in S₂ according to the stoichiometry of the reaction.

Now, we can write the expression of equilibrium constant K_p as follows:

Hence, the equilibrium pressure of H₂ can be calculated using the ICE table as follows:

Hence, the equilibrium pressure of H₂ is 0.0198 atm.