Le Chatelier's Principle Question 5

\ce{CO(g) + 2H2(g) <--> CH3OH(g)} \hskip{2em} \Delta H \lt 0

The synthesis of $\ce{CH3OH(g)}$ from $\ce{CO(g)}$ and $\ce{H2(g)}$ is represented by the equation above. The value of $K_c$ for the reaction at 483 K is 14.5.

Which of the following explains the effect on the equilibrium constant, $K_c$ , when the temperature of the reaction system is increased to 650 K?

K_c

will increase because the activation energy of the forward reaction increases more than that of the reverse reaction.

K_c

will increase because there are more reactant molecules than product molecules.

K_c

will decrease because the reaction is exothermic.

K_c

is constant and will not change.

Approach

Since $\Delta H \lt 0$ , the reaction is exothermic and we can treat heat as a product.

\ce{CO(g) + 2H2(g) <--> CH3OH(g)} + \text{heat}

Since the temperature is increasing, we should expect a shift in equilibrium position to the left since products are being added.

The value of $K_c$ will decrease since more reactants are being made (denominator increasing).

K_c = \frac{[\ce{CH3OH(g)}]}{[\ce{CO(g)}][\ce{H2(g)}]^2}