. For the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ \(K=168\) at \(1273 \mathrm{~K}\). If one starts with \(0.3\) atm of \(\mathrm{CO}_{2}\) and \(12.0 \mathrm{~g}\) of \(\mathrm{C}\) at \(1273 \mathrm{~K}\), will the equilibrium mixture contain (a) mostly \(\mathrm{CO}_{2}\) ? (b) mostly CO? (c) roughly equal amounts of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) ? (d) only C?

When a chemical reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction, it has achieved a dynamic balance referred to as chemical equilibrium. However, even though the visible process appears halted—no change occurs in the concentrations of reactants and products—the molecules continue to react at a constant rate in both directions.

The equilibrium constant, denoted as K, is a number that reflects this ratio of product concentrations to reactant concentrations at equilibrium. For our reaction, \( \mathrm{C}(s) + \mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g) \), the equilibrium constant expression is \( K = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \) where \( [\mathrm{CO}] \) and \( [\mathrm{CO}_2] \) represent the concentrations of carbon monoxide and carbon dioxide in the gas phase, respectively.

In the exercise scenario, an equilibrium constant value of 168 indicates that at 1273 K, the reaction heavily favors the production of CO when equilibrium is achieved. Since carbon in the solid state does not have a concentration in the same way a gas does, it is not included in the expression.

To predict which direction a reaction will proceed to reach equilibrium, chemists use the reaction quotient, Q. This calculation is similar to the equilibrium constant, but with the concentrations or partial pressures of reactants and products at any point before the system has reached equilibrium.

For our reaction, the reaction quotient Q is given by \( Q = \frac{(2x)^2}{0.3-x} \) where x represents the change in concentration of \( \mathrm{CO}_2 \) as the reaction moves toward equilibrium. Initially, Q is zero because CO has not formed yet. Comparing this initial Q value to the equilibrium constant K tells us if a reaction will proceed forward or reverse to reach equilibrium. In our case, since \( Q_{initial} < K \) (0 < 168), we can anticipate the reaction will proceed forward to form more CO and consume \( \mathrm{CO}_2 \) until equilibrium is achieved.

Le Chatelier's Principle provides a qualitative means of predicting the direction of shift in a reaction when outside stresses, such as changes in concentration, pressure, or temperature, are applied to a system at equilibrium. This principle states that if a system at equilibrium experiences a change in conditions, the equilibrium will adjust to counteract the change and re-establish balance.

For instance, if more \( \mathrm{CO}_2 \) were added to our system, Le Chatelier's Principle predicts that the reaction would shift to the right, producing more CO to reduce the additional \( \mathrm{CO}_2 \) in line with the imposed change. Conversely, increasing the concentration of CO would cause the reaction to shift to the left, forming more \( \mathrm{CO}_2 \) and solid carbon. Le Chatelier's Principle helps us understand not only the direction in which reactions will shift but also reinforces the dynamic nature of chemical equilibrium.