Problem 18 - Chemistry The Central Science

Question

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? $$\begin{array}{ll}{\text { (a) } 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)} & {K_{p}=5.0 \times 10^{12}} \\\ {\text { (b) } 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)} & {K_{c}=5.8 \times 10^{-18}}\end{array}$$

Step-by-Step Solution

Answer

Reaction (a) lies to the right, favoring the formation of \(NO_2(g)\) because \(K_p = 5.0 \times 10^{12}\) which is greater than 1. Reaction (b) lies to the left, favoring the formation of \(HBr(g)\) because \(K_c = 5.8 \times 10^{-18}\) which is less than 1.

Step 1: Understand the Equilibrium Constant

The equilibrium constant is a measure of the ratio of concentrations of products to reactants when the reaction has reached equilibrium. For example, in reaction (a), the equilibrium constant can be written as: \[K_p = \frac{[NO_2]^2}{[NO]^2[O_2]}\] If \(K_p > 1\), the reaction lies to the right, favoring the formation of products. If \(K_p < 1\), the reaction lies to the left, favoring the formation of reactants. The same applies for \(K_c\).

Step 2: Analyze Reaction (a)

We are given \(K_p = 5.0 \times 10^{12}\) for the following reaction: \[ 2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g) \] Since \(K_p > 1\), this reaction lies to the right, favoring the formation of products, i.e., \(NO_2(g)\).

Step 3: Analyze Reaction (b)

We are given \(K_c = 5.8 \times 10^{-18}\) for the following reaction: \[ 2 HBr(g) \rightleftharpoons H_2(g) + Br_2(g) \] Since \(K_c < 1\), this reaction lies to the left, favoring the formation of reactants, i.e., \(HBr(g)\).

Step 4: Final Result

From our analysis, we can conclude that: 1. Reaction (a) lies to the right favoring the formation of \(NO_2(g)\). 2. Reaction (b) lies to the left favoring the formation of \(HBr(g)\).

Key Concepts

Chemical Equilibrium

Imagine a tug-of-war where neither side is strong enough to overpower the other; both teams are exerting equal force but are standing still. This scenario is akin to chemical equilibrium in reactions, a state where the rate of the forward reaction (reactants turning into products) equals the rate of the reverse reaction (products reverting back to reactants). At equilibrium, the concentration of reactants and products does not change, even though the reaction is still ongoing. For example, when we consider reaction (a) from our exercise, which has a high equilibrium constant (\( K_p = 5.0 \times 10^{12} \)), it suggests that at equilibrium, the concentration of products (\( \text{NO}_2 \) ) is much higher than the concentration of reactants (\( 2 \text{NO} \) and \( \text{O}_2 \) ). This indicates a product-favored equilibrium state.

In other words, at chemical equilibrium, there's an established balance between the amounts of reactants and products. However, it's important to note that this balance doesn't mean equal concentrations; the equilibrium constant tells us whether reactants or products are favored in the mixture.

Reaction Quotient

When you first mix reactants in a beaker, they may not be at equilibrium. To decipher the reaction's direction, you assess the reaction quotient (\( Q \) ), a 'snapshot' value that compares the concentration of the reactants and products at any point before equilibrium is reached. It's calculated the same way as the equilibrium constant but isn't necessarily equal to it.

Think of \( Q \) as a predictor: if \( Q < K \) , the reaction will proceed forward to create more products. Conversely, if \( Q > K \) , the reaction will go in reverse to produce more reactants. Measuring Q against the equilibrium constant (\( K \) ) informs us about the reaction's progress. For instance, if we had an instantaneous measurement of concentrations in reaction (a), we could calculate \( Q \) and know whether the reaction needs to shift right or left to reach equilibrium.

Le Chatelier's Principle

Ever wonder how a reaction responds to changes in its environment? Le Chatelier's principle provides the answer. It states that if a system at equilibrium is disturbed by a change in concentration, temperature, or pressure, the system will adjust, or shift, to counteract the disturbance and restore equilibrium.

For example, if more \( \text{NO} \) is added to reaction (a), the principle suggests the reaction would shift towards the products, increasing the amount of \( \text{NO}_2 \) to regain balance. Similarly, for reaction (b), if \( \text{H}_2 \) is removed from the system, the reaction will shift to the right to produce more \( \text{H}_2 \) from \( \text{HBr} \). Le Chatelier's principle helps chemists control reactions, predicting how a system will respond to external changes and allowing for manipulation of conditions to maximize product yield.

Dynamic Equilibrium

In a state of dynamic equilibrium, two opposing processes occur at equal rates so that the overall system appears static, but in reality, it is very much in motion. This is not a stagnant state; reactants turn into products and products back into reactants at the same rate.

This process is beautifully balanced – no net change in reactant or product concentrations occurs, providing the illusion of inactivity. An example is reaction (a), where \( 2 \text{NO} \) and \( \text{O}_2 \) continuously combine to form \( 2 \text{NO}_2 \) and vice versa, yet the observable amounts of these substances remain unchanged once equilibrium is reached. Understanding dynamic equilibrium is crucial as it reminds us that chemical equilibrium is not a static but a perpetually adjusting state.