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For the reaction $$ \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ the value of the equilibrium constant at \(825^{\circ} \mathrm{K}\) is \(0.137\). If \(5.0\) moles of \(\mathrm{CO}_{2}, 5.0\) moles of \(\mathrm{H}_{2}, 1.0 \mathrm{~mole}\) of \(\mathrm{CO}\), and \(1.0\) mole of \(\mathrm{H}_{2} \mathrm{O}\) are initially present, what is the composition of the equilibrium mixture ?

Short Answer

Expert verified
The composition of the equilibrium mixture is approximately 3.15 moles of CO2, 3.15 moles of H2, 2.85 moles of CO, and 2.85 moles of H2O.

Step by step solution

01

Write the expression for the reaction quotient (Q)

The reaction quotient (Q) is defined as the ratio of the concentrations of products to reactants, each raised to the stoichiometric coefficient in the balanced chemical equation. For the given reaction: \(CO_{2}(g) + H_{2}(g) \rightleftarrows CO(g) + H_{2}O(g)\) The reaction quotient, Q, is given by: \( Q = \frac{[CO][H_{2}O]}{[CO_{2}][H_{2}]}\) We calculate Q using the initial concentrations to compare it with the equilibrium constant (K) value.
02

Calculate the initial concentrations of the reactants and products

We are given the initial moles of each species, but we need to convert them to concentrations. Assuming the volume of the system is constant, we can relate moles to concentration. Given, initially, - CO2 = 5.0 moles - H2 = 5.0 moles - CO = 1.0 mole - H2O = 1.0 mole To calculate the initial reaction quotient (Q), we can consider the initial mole values as concentrations (as we are comparing them), so: \(Q = \frac{[CO][H_{2}O]}{[CO_{2}][H_{2}]} = \frac{(1.0)(1.0)}{(5.0)(5.0)}= \frac{1}{25}\)
03

Determine the direction in which the reaction will proceed

Now we'll compare the initial reaction quotient (Q) to the equilibrium constant (K) to determine the direction in which the reaction will proceed to achieve equilibrium. Given: \(K = 0.137\) Since, \(Q < K\) (As \(\frac{1}{25}<0.137\)), this means the reaction must proceed in the forward direction (toward products) to reach equilibrium.
04

Set up an ICE (Initial, Change, Equilibrium) table to track the changes of reactants and products

The ICE table helps to track the amounts of reactants and products at each stage: Reactant / Product | Initial | Change | Equilibrium -------------------|---------|---------|------------ CO2 | 5.0 | -x | 5.0 - x H2 | 5.0 | -x | 5.0 - x CO | 1.0 | x | 1.0 + x H2O | 1.0 | x | 1.0 + x Here x represents the amount of reactants consumed and products formed as the reaction proceeds toward equilibrium.
05

Write the equilibrium expression and substitute the equilibrium values

The equilibrium constant (K) expression for the given reaction: \(K = \frac{[CO][H_{2}O]}{[CO_{2}][H_{2}]}\) Insert the equilibrium concentrations from the ICE table: \(0.137 = \frac{(1 + x)(1 + x)}{(5 - x)(5 - x)}\)
06

Solve for x

To find the composition of the equilibrium mixture, we need to solve the equation for x. It is a quadratic equation, but the value of x is small compared to the initial moles. So, we can assume that x is much smaller than 1, 5, or 5 - x. In such approximation: \(0.137 = \frac{x^2}{25}\) Solving for x: \(x^2 = 0.137 \times 25\) \(x^2 = 3.425\) \(x = \sqrt{3.425} \approx 1.85\)
07

Calculate the equilibrium concentrations and complete the ICE table

Now, we'll use the value of x to determine the equilibrium concentrations: Reactant / Product | Initial | Change | Equilibrium -------------------|---------|---------|------------ CO2 | 5.0 | -x | 5.0 - 1.85 H2 | 5.0 | -x | 5.0 - 1.85 CO | 1.0 | x | 1.0 + 1.85 H2O | 1.0 | x | 1.0 + 1.85
08

Write the equilibrium concentrations

The final concentrations at equilibrium are as follows: - CO2: 5.0 - 1.85 ≈ 3.15 moles - H2: 5.0 - 1.85 ≈ 3.15 moles - CO: 1.0 + 1.85 ≈ 2.85 moles - H2O: 1.0 + 1.85 ≈ 2.85 moles The composition of the equilibrium mixture is approximately 3.15 moles of CO2, 3.15 moles of H2, 2.85 moles of CO, and 2.85 moles of H2O.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant (\textbf{K}) is a number that describes the ratio of the concentrations of products to reactants at equilibrium, with each concentration raised to the power of its coefficient in the balanced chemical equation. This value is essential as it indicates the extent of a reaction; a higher value suggests a reaction favors the formation of products, whereas a lower value indicates a reaction that favors reactants.

For example, in the given reaction
\[ CO_2(g) + H_2(g) \rightleftharpoons CO(g) + H_2O(g) \],
the equilibrium constant at \(825 \textdegree C\) is \(0.137\). This means at this temperature, when the reaction has reached equilibrium, the concentration of products (CO and H2O) raised to their stoichiometric coefficients and divided by the concentration of reactants (CO2 and H2) also raised to their stoichiometric coefficients is 0.137.
Reaction Quotient
The reaction quotient (\textbf{Q}) is a critical concept that mirrors the expression for the equilibrium constant but uses the initial concentrations of reactants and products. Calculation of Q is essential to predict the direction in which a reaction will proceed to reach equilibrium.

In the example provided,
\[ Q = \frac{[CO][H_{2}O]}{[CO_{2}][H_{2}]} \],
we calculated Q with the initial concentrations and found that it was less than the equilibrium constant (\textbf{K}). This indicates that the reaction will shift towards the products to achieve equilibrium. Understanding Q helps predict such shifts, which consequently permits adjustments to be made for reactions to reach a desired equilibrium state.
ICE Table
An ICE table—which stands for Initial, Change, and Equilibrium—is a valuable tool used to track the changes in concentration or pressure of reactants and products over the course of a reaction. By organizing the initial amounts, the changes that occur as the system moves towards equilibrium, and the final equilibrium concentrations, an ICE table clarifies how stoichiometry and equilibrium concepts come together.

Using the given reaction as our example, we filled out an ICE table and by assuming x to be much smaller than the initial amounts, we simplified the equilibrium expression leading to a solution for x. Subsequently, we calculated the equilibrium concentrations of all species. This practical approach helps visualize the progression of a reaction and simplifies the process of finding the equilibrium composition.
Le Chatelier's Principle
Le Chatelier's principle is a fundamental rule in chemistry that predicts how a system at equilibrium will respond to disturbances. When a chemical system experiences a change in concentration, pressure, or temperature, the system will adjust to counteract the imposed change and restore a new equilibrium.

Through this principle, we understand that if additional reactants are added, the system will favor the formation of products to reduce reactant concentration, and vice versa. If there is a change in volume or pressure, the system adjusts in the direction that minimizes the change. Similarly, for temperature, an increase will favor the endothermic reaction direction, and a decrease will favor the exothermic direction.

This principle is essential for controlling chemical reactions, informing how to shift equilibria for the desired outcome. Our example doesn't address Le Chatelier's Principle directly, but by understanding it, we can hypothesize how changes in conditions could alter the equilibrium state of the reaction.

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Most popular questions from this chapter

At a certain temperature, an equilibrium mixture of $$ \mathrm{NO}_{2}+\mathrm{SO}_{2} \rightleftarrows \mathrm{NO}+\mathrm{SO}_{3} $$ is analyzed and found to contain the following molar concentrations: \(\left[\mathrm{NO}_{2}\right]=0.100,\left[\mathrm{SO}_{2}\right]=0.300,[\mathrm{NO}]=2.00,\left[\mathrm{SO}_{3}\right]=0.600\) If \(0.500\) moles of \(\mathrm{SO}_{2}\) are introduced at constant temperature, what will be the new concentrations of reactants and products when equilibrium is re-established?

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A solute of formula \(\mathrm{AB}\) is slightly dissociated into \(\mathrm{A}^{+}\) and \(\mathrm{B}^{-}\). In this system, there is a dynamic equilibrium such that \(\mathrm{A}^{+}+\mathrm{B}^{-} \rightleftarrows \mathrm{AB}\). Explain what happens if more acid is introduced into this system.

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