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

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?

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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).

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