Question

For the reaction CO(g) + 3 H2(g) ⇌ H2O(g) + CH4(g), Kc = 189 at 1000 K. If a vessel is filled with these gases such that the initial concentrations are [CO] = 0.0360 M, [H2] = 0.0450 M, [H2O] = 0.0200 M, and [CH4] = 0.0310 M, in which direction will a reaction occur and why?

Answer 1

Answer:

There will be no change since Qc = Kc

Explanation:

Given:

[CO] = 0.0360 M

[H2] = 0.0450 M

[H2O] = 0.0200 M

[CH4] = 0.0310 M

Kc = 189

To determine:

The direction in which the given reaction will occur

Calculation:

The given reaction is:

$CO(g) + 3 H2(g) \rightleftharpoons  H2O(g) + CH4(g)$

The relation between the equilibrium constant Kc and the reaction quotient Qc determines the reaction direction

$Qc = \frac{[Products]}{[Reactants} \\\\For\ the\ given\ reaction\\\\Qc = \frac{[H2O][CH4]}{[CO][H2]^{3} } -----(1)$

If Qc > Kc reaction will shift to the left

If Qc < Kc reaction will shift to the right

If Qc = Kc there will be no shift

Based on equation (1), for the given reaction:

$Qc= \frac{[0.0200][0.0310]}{[0.0360][0.0450]^{3} } =189$

Since Qc=Kc, there will be no change in the reaction direction and the equilibrium will be retained