Consider the reaction: $$\mathrm{NH}_{4} \mathrm{HS}(s) \Longrightarrow \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ At a certain temperature, \(K_{\mathrm{c}}=8.5 \times 10^{-3} .\) A reaction mixture at this temperature containing solid \(\mathrm{NH}_{4} \mathrm{HS}\) has \(\left[\mathrm{NH}_{3}\right]=0.166 \mathrm{M}\) and \(\left[\mathrm{H}_{2} \mathrm{S}\right]=0.166 \mathrm{M} .\) Will more of the solid form, or will some of the existing solid decompose as equilibrium is reached?

The equilibrium constant, denoted as Kc, is fundamental in understanding chemical reactions that reach a state of balance. It describes the ratio of the concentrations of products to reactants at equilibrium for a reaction taking place in a closed system. Mathematically, for a general reaction \( aA + bB \Longrightarrow cC + dD \), the equilibrium constant is expressed as \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \), where the brackets denote the concentrations of the corresponding substances in moles per liter (M).

When dealing with solids, like \(\mathrm{NH}_{4}\mathrm{HS}\) in our exercise, their concentration remains constant and therefore do not appear in the expression for Kc. This greatly simplifies the calculation and interpretation of the equilibrium state. The value of Kc is specific to a particular reaction at a constant temperature, and a higher Kc value implies a greater concentration of products at equilibrium.

The reaction quotient, Q, is a measurement used to determine the direction in which a reaction will proceed to reach equilibrium. It uses the same form as the equilibrium constant Kc, but is calculated using the initial concentrations, not the equilibrium concentrations.

In our case, we calculate Q using the formula \( Q = [\mathrm{NH_3}][\mathrm{H_2S}] \) with the initial concentrations given. The value of Q at any point in the reaction gives us a snapshot of where the system lies in relation to equilibrium. If Q equals Kc, the system is at equilibrium. However, if Q does not equal Kc, the system will shift in the direction that brings Q towards Kc, thus reaching a state of equilibrium.

When a chemical reaction in equilibrium is disturbed, perhaps by changes in concentration, pressure, or temperature, the system will undergo a shift to re-establish equilibrium—a phenomenon described by Le Chatelier's Principle. Depending on the reaction quotient Q and the equilibrium constant Kc, the shift in equilibrium can be predicted.

• If \( Q < K_c \), the reaction shifts to the right, meaning it will produce more products to achieve equilibrium.
• If \( Q > K_c \), the reaction shifts to the left, indicating a production of more reactants.
• If \( Q = K_c \) the reaction is at equilibrium, and no shift will occur.

In the exercise, since Q is greater than Kc, there is a shift towards the reactants, which implies more solid \(\mathrm{NH}_{4}\mathrm{HS}\) is formed.

Chemical reaction calculations are vital to predict the outcomes of reactions under various conditions. This involves calculating the equilibrium constant Kc, determining the reaction quotient Q, and predicting shifts in equilibrium based on the comparison of these values. Through these calculations, one can deduce how a system in disequilibrium will adjust to minimize the disturbance and reach a state of balance once again.

The step-by-step method illustrated in the exercise is an excellent example of such calculations. It emphasizes the importance of first identifying and writing the correct expressions for Kc and Q, and then executing proper comparisons to infer the direction in which the equilibrium will shift. In our exercise, the calculation concluded that more solid \(\mathrm{NH}_{4}\mathrm{HS}\) will form as the system seeks equilibrium.