For the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) \(K_{\text {eq }}=3.59\) at \(900^{\circ} \mathrm{C}\). After the reaction has run for 10 min at \(900^{\circ} \mathrm{C}\), the concentrations are \(\left[\mathrm{CH}_{4}\right]=\) \(1.15 \mathrm{M} ;\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.20 \mathrm{M} ;\left[\mathrm{CS}_{2}\right]=1.51 \mathrm{M} ;\left[\mathrm{H}_{2}\right]=\) \(1.08 \mathrm{M}\). Is this reaction at equilibrium?

Understanding the equilibrium constant (K) is crucial in the study of chemical reactions that can occur in both forward and reverse directions. This constant is a measure of the relative concentrations of products to reactants at equilibrium for a given reaction at a constant temperature.

The general expression for the equilibrium constant is given by:
\[\begin{equation}K = \frac{[products]}{[reactants]}\end{equation}\]
For a reaction, such as \[\begin{equation}aA + bB \rightleftarrows cC + dD\end{equation}\]
, the equilibrium constant expression would be:
\[\begin{equation}K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\end{equation}\]
where the concentrations are raised to the power of their respective coefficients in the balanced equation.

It is also important to note that the equilibrium constant is dimensionless and its value is specific to a particular temperature; changing the temperature will change the value of K. A high K value indicates a greater concentration of products at equilibrium, while a low K implies a greater concentration of reactants.

In the context of our exercise, the given equilibrium constant (\[\begin{equation}K_{eq}=3.59\end{equation}\]) at 900°C, dictates the ratio of the concentration of products to reactants when the reaction system is at equilibrium.

The reaction quotient (Q) is a measure that tells us the direction in which a reaction mixture will proceed to reach equilibrium. It uses the same formula as the equilibrium constant but with the current or initial concentrations of the reactants and products, instead of those at equilibrium.

The calculation for Q allows us to compare its value with the equilibrium constant (K) to predict the shift in the reaction:

• If \[\begin{equation}Q < K\end{equation}\], the forward reaction is favored, and the system will proceed towards products to reach equilibrium.
• If \[\begin{equation}Q > K\end{equation}\], the reverse reaction is favored, and the system will proceed towards reactants.
• If \[\begin{equation}Q = K\end{equation}\], the system is already at equilibrium, and no net change will occur.

In our exercise, we calculated Q and found it to be less than K, which indicates that the reaction will continue to proceed forward to produce more products until Q equals K, signifying that equilibrium has been reached.

When a chemical system at equilibrium is subjected to a change in concentration, temperature, or pressure, Le Chatelier's principle states that the system will adjust to partly counteract the change and re-establish equilibrium.

This principle can be summarized as follows: if a system at equilibrium experiences a disturbance, it responds to minimize the disturbance and return to a state of balance. Adjustments could involve shifting the equilibrium position towards products or reactants, depending on the nature of the change applied.

For example:

• Adding more reactants to the system will shift the equilibrium to produce more products.
• Increasing the temperature of an endothermic reaction will shift the equilibrium towards products.
• Reducing the volume of a gaseous reaction increases pressure, shifting the equilibrium towards the side with fewer moles of gas.

Applying Le Chatelier's principle to the exercise scenario, if we were to alter the concentration of any reactants or products, the system would shift its equilibrium position to counteract our changes until the equilibrium constant (K) value is restored.