91Ó°ÊÓ

An experimenter places the following concentrations of gases in a closed container: \([\mathrm{NOBr}]=7.13 \times 10^{-2} M\) \([\mathrm{NO}]=1.58 \times 10^{-2} M,\left[\mathrm{Br}_{2}\right]=1.29 \times 10^{-2} M .\) These gases then react: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ At the temperature of the reaction, the equilibrium constant \(K_{c}\) is \(3.07 \times 10^{-4} .\) Calculate the reaction quotient, \(Q_{c}\) from the initial concentrations and determine whether the concentration of NOBr increases or decreases as the reaction approaches equilibrium. a.\(Q_{c}=6.33 \times 10^{-4} ;\) the concentration of \(\mathrm{NOBr}\) decreases b.\(Q_{c}=6.33 \times 10^{-4} ;\) the concentration of \(\mathrm{NOBr}\) increases c.\(Q_{c}=1.58 \times 10^{4} ;\) the concentration of \(\mathrm{NOBr}\) increases d.\(Q_{c}=4.65 \times 10^{-4} ;\) the concentration of \(\mathrm{NOBr}\) decreases e.\(Q_{c}=4.65 \times 10^{-4} ;\) the concentration of \(\mathrm{NOBr}\) increases

Step by step solution

01

Determine the Reaction Quotient Formula

The reaction quotient, \( Q_c \), is calculated using the formula for the reaction quotient of the reversible reaction given by: \[ Q_c = \frac{[NO]^2 [Br_2]}{[NOBr]^2} \] where \([NO]\), \([Br_2]\), and \([NOBr]\) are the initial concentrations of the gases.

02

Substitute the Initial Concentrations

Substitute the given initial concentrations into the formula: \([NO] = 1.58 \times 10^{-2} \ M, \; [Br_2] = 1.29 \times 10^{-2} \ M, \; \text{and} \; [NOBr] = 7.13 \times 10^{-2} \ M\): \[ Q_c = \frac{(1.58 \times 10^{-2})^2 (1.29 \times 10^{-2})}{(7.13 \times 10^{-2})^2} \]

03

Calculate the Numerator

Calculate \((1.58 \times 10^{-2})^2 \times (1.29 \times 10^{-2})\): \[ (1.58 \times 10^{-2})^2 = 2.4964 \times 10^{-4} \] and \[ 2.4964 \times 10^{-4} \times 1.29 \times 10^{-2} = 3.220956 \times 10^{-6} \]

04

Calculate the Denominator

Calculate \((7.13 \times 10^{-2})^2\): \[ (7.13 \times 10^{-2})^2 = 5.082169 \times 10^{-3} \]

05

Calculate the Reaction Quotient, \(Q_c\)

Divide the numerator by the denominator to find \(Q_c\): \[ Q_c = \frac{3.220956 \times 10^{-6}}{5.082169 \times 10^{-3}} = 6.33 \times 10^{-4} \]

06

Compare \(Q_c\) with \(K_c\)

The value of \(K_c\) is given as \(3.07 \times 10^{-4}\). Since \(Q_c = 6.33 \times 10^{-4}\) is greater than \(K_c\), the reaction will proceed to the left to reach equilibrium, meaning \([NOBr]\) will increase.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Quotient

The Reaction Quotient, denoted as \( Q_c \), is a valuable tool in chemistry that allows us to determine the current state of a chemical reaction relative to its equilibrium. By comparing \( Q_c \) to the Equilibrium Constant \( K_c \), we can predict the direction in which the reaction must proceed to reach equilibrium.

For the reversible reaction \( 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \), the formula for \( Q_c \) is given by:

• \( Q_c = \frac{[\text{NO}]^2 [\text{Br}_2]}{[\text{NOBr}]^2} \)

Here, \([\text{NO}]\), \([\text{Br}_2]\), and \([\text{NOBr}]\) represent the initial concentrations of the respective gases. It's essential to substitute the initial concentrations given in a problem to calculate \( Q_c \).

In the current problem, the calculation yields a \( Q_c \) of \( 6.33 \times 10^{-4} \), which when compared to \( K_c \) helps in predicting the concentration changes as described in the solution.

Equilibrium Constant

The Equilibrium Constant, \( K_c \), is a crucial concept in chemical equilibrium that describes the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. Each reactant and product concentration is raised to the power of its coefficient in the balanced equation.

For a reversible reaction like \( 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \), \( K_c \) is expressed as:

• \( K_c = \frac{[\text{NO}]^2 [\text{Br}_2]}{[\text{NOBr}]^2} \)

In this problem, the equilibrium constant \( K_c \) is provided as \( 3.07 \times 10^{-4} \). It gives us insight into the favored direction of the reaction at equilibrium.

A value of \( K_c \) smaller than \( Q_c \) indicates that the reaction will shift towards the reactants to reach equilibrium, where more \( \text{NOBr} \) is formed, as explained in the solution.

Concentration Changes in Reactions

Understanding concentration changes in reactions is vital to predict how a system approaches equilibrium. When the Reaction Quotient \( Q_c \) and Equilibrium Constant \( K_c \) are compared, they tell us whether the concentrations of reactants or products will increase or decrease.

In the problems like this one involving \( 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \), observing \( Q_c > K_c \) implies the system shifts left, towards the formation of more reactants.

This means the concentration of \( \text{NOBr} \) will increase as the reaction moves to reach equilibrium. Conversely, if \( Q_c < K_c \), the reaction would shift right, producing more products. This dynamic adjustment of concentrations ensures the system eventually reaches a state where \( Q_c = K_c \), indicating equilibrium.