A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

Short Answer

Expert verified
Using the calculated equilibrium concentrations, Kc for the reaction is found to be 2.12.

Step by step solution

01

Write down initial moles and change in moles

Initially, there are 2.00 mol of CO and 3.00 mol of H2. At equilibrium, 0.478 mol of CH4 is formed. For every mole of CH4 formed, 1 mole of CO reacts and 3 moles of H2 react.
02

Calculate equilibrium moles of each gas

Calculate the moles of the reactants (CO and H2) remaining by subtracting the moles of CH4 formed times their respective ratios from the initial moles. Calculate the moles of the products (CH4 and H2O) by adding the moles of CH4 formed to the initial moles (which is 0 for both products since they are not present at the beginning).
03

Calculate the equilibrium concentrations

Divide the equilibrium moles by the volume of the reaction vessel (10.0 L) to obtain the concentrations of all gases at equilibrium.
04

Write the expression for Kc

Kc is defined as the ratio of the concentration of products to the concentration of reactants, each raised to the power of their coefficients in the balanced chemical equation.
05

Calculate the value of Kc

Substitute the equilibrium concentrations into the Kc expression to get the value of Kc.

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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 by Q, plays a pivotal role in determining how a reaction mixture is related to its equilibrium state. It is mathematically similar to the equilibrium constant, but it applies to systems that may not be at equilibrium. Q is calculated by taking the ratio of the products' concentrations to the reactants' concentrations at any point in time, each raised to the power of their coefficients from the balanced equation.

To calculate Q, you would write an expression identical to the one for the equilibrium constant, Kc, but using the current concentrations of the reacting species. For example, if we consider the reaction given in the exercise, the reaction quotient would be calculated as follows:
\[\begin{equation}Q = \frac{[CH_4][H_2O]}{[CO][H_2]^3}\end{equation}\]
The value of Q can offer insights into which direction the reaction will proceed to reach equilibrium. If Q is less than Kc, the reaction moves forward towards the formation of products. If Q is greater than Kc, the reaction tends to form more reactants.
Equilibrium Constant (Kc)
The equilibrium constant, Kc, is a number that characterizes the chemical equilibrium of a reaction. Its value is constant for a given reaction at a constant temperature. For the reaction in our exercise, Kc is defined based on the concentrations of gases at equilibrium. The expression for Kc would be structured similarly to the reaction quotient:
\[\begin{equation}K_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3}\end{equation}\]
However, Kc specifically uses the equilibrium concentrations. By comparing different Kc values for the same reaction at different temperatures, we can understand how the equilibrium position changes with temperature. Calculating Kc accurately requires precise measurements of the equilibrium concentrations, and from this, we can predict the extent of the reaction and make qualitative assessments about the system.
Le Chatelier's Principle
Le Chatelier's Principle provides a qualitative prediction of how a system at equilibrium responds to changes in concentration, pressure, volume, or temperature. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle helps us understand how to alter conditions to favor the formation of products or reactants.

For instance, if we increase the concentration of a reactant, the system shifts to the right, forming more products. If we remove a product, the equilibrium also shifts to the right to produce more product and re-establish equilibrium. When pressure is applied to a gas reaction with unequal moles on either side, the equilibrium shifts towards the side with fewer moles of gas. Temperature changes effectively change the value of Kc, shifting the equilibrium towards endothermic or exothermic direction depending on whether heat is added or removed.
Equilibrium Concentrations
Equilibrium concentrations refer to the concentration of each species in a reaction mixture when the system has reached a state of equilibrium. These concentrations remain constant over time as the rates of the forward and reverse reactions are equal. In practice, calculating these concentrations typically involves initial concentration measurements, understanding the stoichiometry of the reaction, and applying the equilibrium constant expression.

In our exercise, we found the equilibrium concentrations by using the initial number of moles of reactants, the change in moles as the reaction progressed, and the volume of the reaction vessel. By substituting these values into the Kc expression, you can solve for unknown concentrations. Equilibrium concentrations are essential for calculating Kc and for predicting how a system will respond to changes using Le Chatelier's principle.

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Most popular questions from this chapter

At \(500^{\circ} \mathrm{C}, K_{c}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\). If analysis shows that the composition is \(3.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 2.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\) and \(0.500 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\), is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Consider the equilibrium \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). (a) If the partial pressure of \(\mathrm{CO}_{2}\) is increased, what happens to the partial pressure of \(\mathrm{CH}_{4}\) ? (b) If the partial pressure of \(\mathrm{CH}_{4}\) is decreased, what happens to the partial pressure of \(\mathrm{CO}_{2}\) ? (c) If the concentration of \(\mathrm{CH}_{4}\) is increased, what happens to the equilibrium constant for the reaction? (d) If the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is decreased, what happens to the concentration of \(\mathrm{CO}_{2}\) ?

Calculate the standard reaction free energy of each of the following reactions: (a) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), K=41\) ar \(400 \mathrm{~K}\) (b) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.0 \times 10^{4}\) at \(700 \mathrm{~K}\)

At \(2500 \mathrm{~K}, K_{e}=20\) for the reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}(\mathrm{g}) .\) An analysis of a reaction vessel at \(2500 \mathrm{~K}\) revealed the presence of \(0.18 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{Cl}_{2}, 0.31 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~F}_{2}\), and \(0.92 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{ClF}\). Will \(\mathrm{ClF}\) tend to form or to decompose as the reaction proceeds toward cquilibrium?

Explain why the following cquilibria are heterogeneous and write the reaction quotient \(Q\) for each one. (a) \(2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{FcCl}_{3}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{KNO}_{3}\) (s) \(\rightleftharpoons 2 \mathrm{KNO}_{2}\) (s) \(+\mathrm{O}_{2}\) (g)

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