Problem 30 - Chemistry and Chemical Reactivity

Question

The equilibrium constant for the reaction $$ \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g}) $$ is \(1.7 \times 10^{-3}\) at \(2300 K\) (a) What is \(K\) for the reaction when written as follows? $$ \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g}) $$ (b) What is \(K\) for the following reaction? $$ 2 \mathrm{NO}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

(a) \( K \approx 0.0412 \). (b) \( K \approx 588.24 \).

Step 1: Determine the reaction for part (a)

The original reaction is \( \mathrm{N}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g}) \) with an equilibrium constant \( K = 1.7 \times 10^{-3} \). For part (a), the reaction is \( \frac{1}{2} \mathrm{N}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g}) \). This reaction is obtained by taking the square root of the original reaction.

Step 2: Adjust the equilibrium constant for part (a)

When a reaction is modified so that the stoichiometric coefficients are divided by \( n \), the equilibrium constant for the new reaction is the original \( K \) raised to the power of \( \frac{1}{n} \). Here, \( n = 2 \), so we calculate \( K' = (1.7 \times 10^{-3})^{\frac{1}{2}} \).

Step 3: Calculate the new equilibrium constant for part (a)

Calculate \( K' = (1.7 \times 10^{-3})^{\frac{1}{2}} = \sqrt{1.7 \times 10^{-3}} \approx \sqrt{0.0017} \approx 0.0412 \). Hence, the equilibrium constant \( K' \) for the reaction \( \frac{1}{2} \mathrm{N}_2 + \frac{1}{2} \mathrm{O}_2 \rightleftarrows \mathrm{NO} \) is approximately 0.0412.

Step 4: Determine the reaction for part (b)

For part (b), the reaction is the reverse of the original reaction: \( 2 \mathrm{NO}(\mathrm{g}) \rightleftarrows \mathrm{N}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \). This means we need to take the reciprocal of the original equilibrium constant \( K \).

Step 5: Calculate the equilibrium constant for part (b)

For the reverse reaction, the new equilibrium constant \( K'' = \frac{1}{K} \). Thus, \( K'' = \frac{1}{1.7 \times 10^{-3}} \approx 588.24 \). Therefore, the equilibrium constant \( K'' \) for the reaction \( 2 \mathrm{NO} \rightleftarrows \mathrm{N}_2 + \mathrm{O}_2 \) is approximately 588.24.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a crucial concept in understanding chemical equilibria. It gives us a ratio of product concentrations to reactant concentrations raised to the power of their stoichiometric coefficients at equilibrium. This constant is specific to a particular reaction at a given temperature.
When we have a reaction, such as \( \mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftarrows 2\mathrm{NO}(\mathrm{g}) \), the equilibrium constant is defined as:\[K = \frac{[NO]^2}{[N_2][O_2]}\]It's important to emphasize that \(K\) values are only affected by temperature changes. In examples where the stoichiometric coefficients are altered, as in part (a) of the problem, the value must be adjusted by raising \(K\) to a power equal to the reciprocal of the stoichiometric change. This versatility in adjusting \(K\) is pivotal when considering different stoichiometries of the same reaction.

Reversible Reactions

Reversible reactions are those chemical reactions where the reactants convert to products and, simultaneously, the products can convert back to reactants. This can be symbolized by the double arrow, \( \rightleftarrows \), showing that the reaction can proceed in both directions.
In the context of chemical equilibrium, a reversible reaction may achieve a state where the rates of the forward and reverse reactions are equal, dictating that no net change in concentrations takes place. For example, in the exercise, both reactions \( \mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftarrows 2\mathrm{NO}(\mathrm{g}) \) and its reverse \( 2\mathrm{NO}(\mathrm{g}) \rightleftarrows \mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \) are balanced in this state of equilibrium.
Understanding reversible reactions is essential for correctly applying the concept of equilibrium constants, which are framed around these two-way processes.

Stoichiometric Coefficients

Stoichiometric coefficients are numbers placed before the compounds in a chemical equation to balance and represent the ratio in which reactants and products participate in the reaction. They have a significant effect on calculations involving the equilibrium constant.
In our exercise, the equation \( \mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftarrows 2\mathrm{NO}(\mathrm{g}) \) has stoichiometric coefficients of 1, 1, and 2, respectively. When modifying the reaction, as in part (a), by dividing the coefficients, it becomes \( \frac{1}{2} \mathrm{N}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g}) \). This requires altering the equilibrium constant by taking the square root, thus adjusting \(K\) to reflect the new stoichiometry. This change is captured mathematically as raising the equilibrium constant to the power of the change applied to the reaction coefficients.

Reaction Quotients

The reaction quotient, denoted by \(Q\), is similar to the equilibrium constant but can be calculated at any point during the reaction, not just at equilibrium. It is given by the same expression as \(K\), involving the concentrations of reactants and products raised to their respective stoichiometric coefficients.
When \(Q\) equals \(K\), the system is at equilibrium. If \(Q < K\), the forward reaction is favored, and the system proceeds to produce more products. Conversely, if \(Q > K\), the reverse reaction is favored, indicating that the reaction will proceed to form more reactants. This relationship provides an insight into the reaction's current position relative to its equilibrium state.
Understanding the reaction quotient is essential for predicting how a system will respond to changes in concentration, pressure, or temperature, aligning the theoretical computations of equilibrium constants with observable chemical behavior.