Problem 14 - Chemical Principles: The Quest for Insight

Question

If \(Q=1.0 \times 10^{50}\) for the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants, or will it be at equilibrium?

Step-by-Step Solution

Answer

With a reaction quotient, Q, of 1.0 x 10^50, this reaction would have a tendency to form more reactants to reach equilibrium.

Step 1: Understand the Reaction Quotient

The reaction quotient, Q, for the given reaction compares the concentrations (or partial pressures) of the products to the reactants. It is calculated based on the expression Q = [CO2]/([C][O2]), where the concentrations of solid carbon (C) and gas oxygen (O2) will affect the value of Q.

Step 2: Compare Q to the Equilibrium Constant

Compare the value of Q to the equilibrium constant K for the reaction at 25°C. If Q is much greater than K, the reaction will proceed in the reverse direction (forming more reactants). If Q is smaller than K, the reaction will proceed in the forward direction (forming more products). If Q is equal to K, the reaction is at equilibrium.

Step 3: Determine the Reaction Tendency

Since no K value is provided, we assume that a Q value as large as 1.0 x 10^50 is extremely high, suggesting the reaction has proceeded far to the right (towards the products). This likely exceeds the K value at 25°C and suggests that the reaction will tend to form more reactants (move in the reverse direction) until equilibrium is reached.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. It is important to understand that equilibrium does not necessarily mean the reactants and products are present in equal proportions, but rather that their concentrations have stabilized at a particular ratio.

In the context of the exercise, the concept of chemical equilibrium is used to determine whether the reaction will form more reactants or products, or if it will remain unchanged. The understanding of the reaction quotient (Q) plays a key role in predicting the direction of the shift in equilibrium if the system is currently not in equilibrium.

Equilibrium Constant K

The equilibrium constant (K) is a value that expresses the ratio of concentrations of products to reactants at equilibrium at a given temperature. It is specific for a particular reaction at a specified temperature. Mathematically, for a reaction aA + bB \rightleftharpoons cC + dD, the equilibrium constant K is given by
\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
where the concentration of each species is raised to the power of its coefficient in the balanced chemical equation.

When comparing the reaction quotient Q to the equilibrium constant K, it's possible to predict the direction in which the reaction needs to move to reach equilibrium. If the reaction quotient Q is greater than K, as suggested by the high value given in the exercise, the reaction would tend to move in the reverse direction to reach equilibrium, decreasing the Q value.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle helps us predict the effect of changes in concentration, pressure, and temperature on the position of equilibrium in a chemical reaction.

Applying this principle to the given exercise, if a reaction has a Q value far different from K, the system is not at equilibrium and will shift to restore it. A Q value significantly higher than the K value for the reaction implies that there is a large concentration of products, and so the reaction should shift to form more reactants according to Le Chatelier's Principle, moving the reaction quotient back towards the value of the equilibrium constant.