Problem 23 - Chemistry The Central Science

Question

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at 1000 \(\mathrm{K}\) . (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2},\) or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) (c) Calculate \(K_{c}\) for \(\operatorname{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\)

Step-by-Step Solution

Answer

(a) At 1000 K, the equilibrium favors the reactants NO and Br2 since Kc < 1. (b) The Kc for the reverse reaction is 76.9. (c) The Kc for the new reaction with halved product coefficient is 0.114.

Step 1: (a) Determine which side is favored in the equilibrium

First, let's look at the given reaction and its equilibrium constant: \[2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\] \[K_{c} = 1.3 \times 10^{-2}\] The equilibrium constant, Kc, represents the ratio of the concentrations of the products to the reactants. If Kc >> 1, products are favored, and if Kc << 1, reactants are favored. Here, Kc is smaller than 1, so the equilibrium will favor the reactants, which are NO and Br2.

Step 2: (b) Calculate Kc for the reverse reaction

Given that Kc for the original reaction is 1.3 x 10^-2, we will find the Kc for the reverse reaction: \[2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\] The relationship between the equilibrium constants of forward and reverse reactions is given by: \[K_c^{'(reverse)} = \frac{1}{K_c^{(forward)}}\] Substitute the given value: \[K_c^{(reverse)} = \frac{1}{1.3 \times 10^{-2}}\] Now, calculate Kc for the reverse reaction: \[K_c^{(reverse)} = 76.9\]

Step 3: (c) Calculate Kc for the new reaction with halved product coefficient

The new reaction is: \[\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\] First, let's rewrite the original reaction by dividing each coefficient by 2: \[\mathrm{NO}(g) + \frac{1}{2}\mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{NOBr}(g)\] Now, to find the Kc for this new reaction, we can use the formula: \[K_c^{(new)} = \sqrt[K_c^{(original)}]{n}\] where n is the new coefficient and Kc (original) is the given Kc for the original reaction. Here, n = 2 (this is because the new reaction is created by halving the coefficients of all species in the initial reaction). \[K_c^{(new)} = \sqrt[1.3 \times 10^{-2}]{2}\] Now, calculate Kc for the new reaction: \[K_c^{(new)} = 0.114\] So, the equilibrium constants for the reactions are: (a) Kc (original) = 1.3 x 10^-2 (favoring NO and Br2) (b) Kc (reverse) = 76.9 (c) Kc (new) = 0.114

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in no overall change in the concentration of reactants and products over time. In a chemical equation, we depict this by the double arrows pointing in opposite directions, as shown in the reaction
\(2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g)\).
at a certain temperature. At equilibrium, the concentrations remain constant but are not necessarily equal.

An equilibrium constant (\(K_c\)) value gives us an insight into the composition of the equilibrium mixture. A low \(K_c\) value (much less than 1) signifies that, upon reaching equilibrium, the reactants are favored, meaning their concentrations will remain comparatively high, while high \(K_c\) values (much greater than 1) indicate a product-favored equilibrium, where the products have a higher concentration. In our exercise, a \(K_c = 1.3 \times 10^{-2}\) tells us that at 1000 K, the reactants, NO and \(\text{Br}_2\), are favored over the product NOBr.

Reaction Quotient

The reaction quotient (\(Q\)) is a measure that tells us the direction in which a reaction is likely to proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant \(K_c\), but with the initial concentrations of reactants and products rather than their equilibrium concentrations.

When analyzing the reaction quotient, we consider:

• If \(Q = K_c\), the system is at equilibrium.
• If \(Q < K_c\), the reaction proceeds forward to produce more products until equilibrium is reached.
• If \(Q > K_c\), the reaction proceeds in reverse to produce more reactants until equilibrium is established.

In the textbook exercise provided, the reaction quotient is not explicitly calculated, but understanding its role helps us appreciate how the system shifts to reach equilibrium. If we had initial concentrations, we could compare \(Q\) to the provided \(K_c\) to predict the direction of the reaction.

Le Chatelier's Principle

Le Chatelier's Principle is a qualitative tool that predicts how a change in conditions (such as concentration, pressure, or temperature) can affect the position of equilibrium. Essentially, if a system at equilibrium is disturbed by a change in conditions, the system responds by adjusting in a way that counteracts the change and re-establishes equilibrium.

This principle can be understood through different changes:

• Increasing the concentration of reactants will shift the equilibrium to the right, favoring product formation,
• Increasing the concentration of products shifts it to the left, favoring reactants,
• Changes in pressure for gaseous reactions also shift the equilibrium depending on the mole change in the reaction, and
• Temperature effects depend on whether the reaction is exothermic or endothermic.

For instance, if the concentration of \(NO\) or \(\text{Br}_2\) is increased for the given reaction, Le Chatelier's Principle predicts a shift toward the right to form more NOBr, counteracting the concentration change. Such a shift is indicated without altering the constants (\(K_c\) and \(Q\)), which only change with temperature.