The reversible reaction between hydrogen chloride gas and one mole of oxygen gas produces steam and chlorine gas: $$4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=0.79$$ Predict the direction in which the system will move to reach equilibrium if one starts with (a) \(P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{HCl}}=P_{\mathrm{O}_{2}}=0.20 \mathrm{~atm}\) (b) \(P_{\mathrm{HCl}}=0.30 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.35 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.15 \mathrm{~atm}\)

The equilibrium constant (K) is fundamental to understanding chemical equilibrium in reversible reactions. It quantifies the ratio of product concentrations (or partial pressures for gases) to reactant concentrations at equilibrium, raised respectively to their coefficients in the balanced chemical equation. For the provided reaction, where hydrogen chloride gas and oxygen gas produce chlorine gas and steam, the equilibrium constant is denoted as:

\[ K = \frac{(P_{{Cl}_2})^2 (P_{{H}_2O})^2}{(P_{HCl})^4 (P_{O_2})} \]
The numerical value of K (0.79 in this case) indicates the extent to which a reaction proceeds before reaching equilibrium. A lower K value suggests a reaction favors the reactants, while a higher K indicates a product-favored reaction. When a system is at equilibrium, the concentrations of reactants and products remain constant over time, balancing each other out, and the rate of the forward reaction equals the rate of the reverse reaction.

The reaction quotient (Q) is very similar to the equilibrium constant but is crucial for determining the direction in which an unsettled system will move to reach equilibrium. Calculated with the same formula as K, Q uses the initial concentrations or partial pressures instead of those at equilibrium.

In the given scenarios, we calculate Q using the provided partial pressures:
\[ Q = \frac{(P_{{Cl}_2})^2 (P_{{H}_2O})^2}{(P_{HCl})^4 (P_{O_2})} \]
Comparing Q to K tells us whether the system will shift to the right, favoring products (if \(Q < K\)), or to the left, favoring reactants (if \(Q > K\)). If \(Q = K\), the system is already at equilibrium. This comparison predicts the reaction's shift to restore equilibrium, showcasing the reaction quotient's role as a 'snapshot' of the system's current state.

Partial pressures pertain to the individual pressure contributions by each gas in a mixture. According to Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressures of each component gas.

The exercise uses partial pressures to calculate both the equilibrium constant and the reaction quotient. In gas-phase reactions, partial pressures are used instead of concentrations because gases mix uniformly, and their pressures are directly proportional to their molar amounts at a given temperature and volume. The calculation of Q for the problem clearly illustrates using partial pressures to predict the direction of a reaction shift, which is integral to understanding how conditions can change in a gas mixture not yet at equilibrium.

Le Chatelier's Principle provides a qualitative insight into how a system at equilibrium responds to changes in concentration, pressure, or temperature. It states that if an external stress is applied to a system at equilibrium, the system will adjust by shifting its equilibrium position to counteract the effect of the disturbance.

In the context of the exercise, where all initial partial pressures are the same (scenario a) or varied (scenario b), Le Chatelier's Principle helps explain how the system will react. When Q exceeds K, the system is stressed with an excess of products, and according to this principle, it will shift towards the reactants to reduce the product concentration. The same principle applies when there’s a change in gas pressure or volume—affecting partial pressures—and the reaction equilibrium will shift accordingly in the direction that counterbalances the change.