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The reaction quotient, denoted as \( Q \), is a useful concept in the study of chemical reactions. It helps us determine the progress of a reaction toward equilibrium. For any given chemical reaction, \( Q \) is calculated using the concentrations of the products and reactants at a specific moment in time, not necessarily at equilibrium.
To calculate \( Q \), you use the formula:

• Place the concentrations of the products with each raised to the power of their stoichiometric coefficients in the numerator.
• Place the concentrations of the reactants with each raised to the power of their stoichiometric coefficients in the denominator.

In our case for the ammonia synthesis reaction, \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \), \( Q \) is represented as:
\[Q = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}\]
This equation allows comparison with the equilibrium constant to predict the direction in which the reaction will proceed to achieve equilibrium.

The equilibrium constant, \( K \), is a fundamental concept in chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium, each raised to their respective stoichiometric coefficients.
For any reaction at equilibrium, \( K \) remains constant as long as the temperature does not change. Unlike \( Q \), \( K \) specifically describes the position of equilibrium.
In the context of the ammonia synthesis reaction, the equilibrium constant \( K_p \) is:\[K_p = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}\]
At \( 650 \mathrm{K} \), \( K_{p} = 4.3 \times 10^{-4} \).
This value tells us the ratio at which hydrogen and nitrogen produce ammonia when the system has reached equilibrium. By comparing \( Q \) and \( K \), we discern whether the reaction will move forward, reverse, or if it鈥檚 already at equilibrium.

The ammonia synthesis reaction is a classic example of chemical equilibrium in action. It involves the combination of nitrogen gas, \( N_2 \), and hydrogen gas, \( H_2 \), to form ammonia, \( NH_3 \). This is represented by the balanced chemical equation:
\[N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\]
Being a reversible reaction, it can reach an equilibrium where the rate of formation of ammonia equals the rate of its decomposition back into nitrogen and hydrogen.

Several factors, such as temperature, pressure, and concentrations, influence this balance. In the synthesis of ammonia, conditions are often adjusted to favor the forward reaction to produce more ammonia. When comparing \( Q \) with \( K_p \), if \( Q < K_p \), the forward reaction is favored, indicating that more product will form until equilibrium is achieved. This is exactly the situation in our exercise at \( 650 \mathrm{K} \), where more ammonia will form until equilibrium is reached.