Question

A gaseous mixture consisting of \(2.23 \mathrm{mmol} \mathrm{N}\) and \(6.69 \mathrm{mmol}\) \(\mathrm{H}_{2}\) in a \(500 . \mathrm{mL}\) container was heated to \(600 . \mathrm{K}\) and allowed to reach equilibrium. Will more ammonia be formed if that equilibrium mixture is then heated to 700 . \(\mathrm{K}\) ? For \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g}), K=1.7 \times 10^{-3}\) at \(600 . \mathrm{K}\) and \(7.8 \times 10^{-5}\) at \(700 . \mathrm{K}\).

Short answer

No, more ammonia will not be formed; instead, the amount of ammonia will decrease if the equilibrium mixture is heated to 700 K.

Step-by-step solution

Determine the Reaction Quotient

We must first calculate the reaction quotient (Q) to predict the direction the reaction will shift when heated to 700 K. Q is calculated by determining the partial pressures of the reactants and products. However, we do not have direct information about the concentrations at equilibrium; therefore, we cannot calculate Q immediately.

Analyze the Effect of Temperature on Equilibrium

Considering the Van't Hoff equation and Le Chatelier's Principle, an increase in temperature for an exothermic reaction (formation of NH3 from N2 and H2) will cause the equilibrium to shift towards the reactants to absorb the extra heat. Since K decreases with an increase in temperature from 600 K to 700 K, we can conclude that more reactants (N2 and H2) and less product (NH3) will be favored at the higher temperature.

Predict the Direction of Shift

Given that the equilibrium constant (K) at 600 K is larger than K at 700 K, heating the mixture will result in a decrease in the concentration of NH3 because the reaction will shift to the left to re-establish equilibrium at the new temperature, which has a smaller K value.

Key concepts

Van't Hoff Equation

The Van't Hoff equation is a powerful tool for understanding how changes in temperature affect the position of chemical equilibrium. The equation is represented as follows:

\[\begin{equation}ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\end{equation}\]
where \(K_1\) and \(K_2\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\) respectively, \(R\) is the universal gas constant, and \(\Delta H^\circ\) is the standard enthalpy change of the reaction. If the reaction is exothermic (negative \(\Delta H^\circ\)), increasing the temperature will cause the equilibrium constant to decrease, indicating a shift in equilibrium toward the reactants. Conversely, for an endothermic reaction (positive \(\Delta H^\circ\)), an increase in temperature increases the value of K, shifting equilibrium towards the products. This explains why the equilibrium constant for the production of ammonia decreases as the temperature rises from 600 K to 700 K, as given in the original exercise.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. In the context of our exercise, heating the equilibrium mixture of nitrogen, hydrogen, and ammonia amounts to adding heat to the system. According to Le Chatelier's Principle, the system will react by favoring the endothermic direction of the reaction, in this case, the decomposition of ammonia into nitrogen and hydrogen, to absorb the excess heat. Since the formation of ammonia is exothermic, increasing the temperature will result in less ammonia being formed, and more nitrogen and hydrogen. By applying this principle, we can predict that heating to 700 K will cause the equilibrium to shift towards the reactants, agreeing with the conclusion drawn from the Van't Hoff equation that more ammonia will not be formed at the higher temperature.

Reaction Quotient (Q)

The reaction quotient (Q) is a measure of the relative amounts of products and reactants present during a reaction at a given moment, and it is essential for predicting the direction in which a reaction will proceed to reach equilibrium. It uses the same expression as the equilibrium constant (K), but with the current concentrations or partial pressures instead of those at equilibrium.

\[\begin{equation}Q = \frac{[products]}{[reactants]}^{stoichiometric ~coefficients}\end{equation}\]
In the initial state of the exercise, because we lack information about the actual concentrations of reactants and products at equilibrium, we cannot calculate Q directly. Typically, if \(Q < K\), the system will shift towards the products to reach equilibrium; if \(Q > K\), the system will shift towards the reactants. This quantifiable relationship is invaluable for understanding the direction of shift when conditions such as temperature change.

Equilibrium Constant (K)

The equilibrium constant (K) is a numerical value that indicates the ratio of concentrations of products to reactants at equilibrium for a reversible chemical reaction at a constant temperature. For the reaction in our exercise, the equilibrium constant expression is

\[\begin{equation}K = \frac{[NH_3]^2}{[N_2][H_2]^3}\end{equation}\]
where the brackets represent the concentration of the gases in moles per liter. The value of K is a measure of the extent of the reaction; a large K indicates a greater concentration of products, while a small K favors reactants. Between the temperatures given in the exercise, K decreases significantly, indicating that the concentration of ammonia at equilibrium will be lower at 700 K compared to 600 K. This equilibrium constant serves as the benchmark used in conjunction with Q to determine reaction shifts under changing conditions.