Problem 83
Question
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 \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).
Step-by-Step Solution
Verified Answer
Using the calculated equilibrium concentrations, Kc for the reaction is found to be 2.12.
1Step 1: 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.
2Step 2: 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).
3Step 3: 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.
4Step 4: 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.
5Step 5: Calculate the value of Kc
Substitute the equilibrium concentrations into the Kc expression to get the value of Kc.
Key Concepts
Reaction QuotientEquilibrium Constant (Kc)Le Chatelier's PrincipleEquilibrium Concentrations
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.
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.
\[\begin{equation}K_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3}\end{equation}\]
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.
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.
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.
Other exercises in this chapter
Problem 80
At \(2500 \mathrm{~K}, K_{e}=20\) for the reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}(\mathrm{g}) .\)
View solution Problem 82
The equilibrium constant \(K=3.5 \times 10^{4}\) for the reaction \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})=\mathrm{PCl}_{5}(\mathrm{~g})\) a
View solution Problem 84
A mixture consisting of \(1.000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and \(1.000 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is placed in a \(10.00
View solution Problem 87
A mixture of \(0.0560 \mathrm{~mol} \mathrm{O}_{2}\) and \(0.0200 \mathrm{~mol}\) \(\mathrm{N}_{2} \mathrm{O}\) is placed in a \(1.00-\mathrm{L}\) reaction vess
View solution