Problem 37 - Chemistry and Chemical Reactivity

Question

Which of the following correctly relates the two equilibrium constants for the two reactions shown? $$\begin{array}{cc} \operatorname{NOCl}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) & K_{1} \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NOCl}(\mathrm{g}) & K \end{array}$$ (a) \(K_{2}=-K_{1}^{2}\) (c) \(K_{2}=1 / K_{1}^{2}\) (b) \(K_{2}=1 /\left(K_{1}\right)^{1 / 2}\) (d) \(K_{2}=2 K_{1}\)

Step-by-Step Solution

Answer

Option (c), \( K_2 = \frac{1}{K_1^2} \), is correct.

Step 1: Analyze each reaction

First, break down the two given reactions and their respective equilibrium constants. **Reaction 1:** \[ \text{NOCl}(g) \rightleftharpoons \text{NO}(g) + \frac{1}{2} \text{Cl}_2(g) \] with equilibrium constant \( K_1 \).**Reaction 2:** \[ 2\text{NO}(g) + \text{Cl}_2(g) \rightleftharpoons 2 \text{NOCl}(g) \]with equilibrium constant \( K_2 \).

Step 2: Express K2 in terms of K1

We need to find a relationship between the equilibrium constants. Reaction 2 can be found by reversing Reaction 1, and then multiplying by 2.- Reversing Reaction 1 gives: \[ \text{NO}(g) + \frac{1}{2}\text{Cl}_2(g) \rightleftharpoons \text{NOCl}(g) \] - The equilibrium constant for this reversed reaction is \( \frac{1}{K_1} \).- Multiplying the reversed reaction by 2 gives Reaction 2: \[ 2\text{NO}(g) + \text{Cl}_2(g) \rightleftharpoons 2\text{NOCl}(g) \] - When multiplying a reaction, the equilibrium constant becomes the original constant raised to the power of the factor by which the equation is multiplied (2 in this case), therefore, \[ K_2 = \left(\frac{1}{K_1}\right)^2 = \frac{1}{K_1^2} \]

Step 3: Compare calculated K2 with options

Now, compare \( K_2 = \frac{1}{K_1^2} \) with the provided options:(a) \( K_2 = -K_1^2 \)(b) \( K_2 = 1 /(K_1)^{1/2} \)(c) \( K_2 = 1 / K_1^2 \)(d) \( K_2 = 2K_1 \)The correct relationship is option (c), \( K_2 = \frac{1}{K_1^2} \).

Key Concepts

Reversible Reactions

In chemistry, reactions aren’t always one-way streets. Some reactions can go both ways, which we call reversible reactions. These are special because they don’t just proceed until all reactants are gone. Instead, they can reach a state where reactants and products are in a balance, constantly converting back and forth. This back and forth process happens until a point is reached where the rates of the forward and reverse reactions are equal.

If you think of reversible reactions like a game of ping-pong, the reactants and products are the ball moving back and forth over the net. The game doesn’t stop, but at equilibrium, the ball is exchanged equally at both sides. In our exercise, we saw reversible reactions involving NOCl converting to NO and Cl₂ and vice versa.

Remember, the reversibility of a reaction is not just about going in reverse but reaching a certain balance, which leads us to the concept of chemical equilibrium.

Chemical Equilibrium

Chemical equilibrium is when a reversible reaction stops changing in concentration. Imagine you're riding a seesaw; when both sides hold the same weight, the seesaw is balanced. Similarly, in chemical reactions, equilibrium is when the forward reaction occurs at the same rate as the backward reaction. This doesn't mean the reactants and products are equal in quantity, but their rate of change is the same.

At this point, no visible changes occur, but molecular activity continues. Chemical equilibrium is dynamic, meaning molecular reactions still occur but without noticeable shifts in concentration. In the exercise, when understanding the equilibrium constants (K), they reflect this state of balance. A higher equilibrium constant indicates that, at equilibrium, the products are favored more than the reactants, and vice versa.

Understanding equilibrium is central to predicting how reactions respond to changes, such as pressure or temperature changes, known as Le Chatelier's Principle, but that's a story for another time.

Reaction Quotients

The reaction quotient is like a snapshot of a reaction's progress at a given moment. It's a way to see if a reaction is at equilibrium by comparing it to the equilibrium constant, K. Calculated similarly to K, the reaction quotient (Q) uses the current concentrations of the reactants and products. If Q equals K, the system is at equilibrium. If not, the reaction will proceed in the direction that leads Q to equal K.

In the context of our exercise, considering the reaction quotient is essential for understanding how close a reaction is to reaching equilibrium. It provides a way to predict the direction a reaction will shift, driving either forward or in reverse to reach that balanced state.

Think of Q as a guiding compass. It helps chemists understand current conditions and predict how reactions behave under change. A key takeaway is how critical the reaction quotient is as a tool to assess and guide reactions toward equilibrium.