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The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.?
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The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2...
Given:
Equilibrium constant (Kp) = 3.0
Initial partial pressure of CO = pCO = 0.48 bar
Initial partial pressure of CO2 = pCO2 = 0 bar
To Find:
Equilibrium partial pressures of CO and CO2
Explanation:
1. Writing the Equilibrium Expression:
The given reaction is CO2(g) + C(s) ⇌ 2CO(g)
The equilibrium constant expression for this reaction is given by Kp = (pCO)^2 / pCO2
2. Setting up the Equation:
We are given that Kp = 3.0. Substituting this value into the equilibrium constant expression, we get:
3.0 = (pCO)^2 / pCO2
3. Solving for the Equilibrium Partial Pressures:
Let's assume the equilibrium partial pressures of CO and CO2 to be x and y respectively.
According to the equation, we can write:
3.0 = x^2 / y
4. Using the Initial Partial Pressures:
At the start of the reaction, the partial pressure of CO is 0.48 bar and the partial pressure of CO2 is 0 bar.
Substituting these values into the equation, we get:
3.0 = (0.48)^2 / 0
5. Analyzing the Equation:
The equation shows that the denominator is zero, which means that the partial pressure of CO2 is zero at the start. This indicates that the reaction has not yet occurred.
6. Reaction Direction:
To determine the direction in which the reaction will proceed, we need to compare the initial partial pressures with the equilibrium partial pressures.
Since the initial partial pressure of CO2 is zero and the equilibrium constant expression contains pCO2 in the denominator, the reaction will proceed in the forward direction, converting CO2 into CO.
7. Calculating the Equilibrium Partial Pressures:
To find the equilibrium partial pressures of CO and CO2, we need to solve the equation:
3.0 = x^2 / y
8. Conclusion:
Since the reaction proceeds in the forward direction, the equilibrium partial pressure of CO2 will be zero, and the equilibrium partial pressure of CO can be calculated by solving the equation.
Equilibrium constant (Kp) = 3.0
Initial partial pressure of CO = pCO = 0.48 bar
Initial partial pressure of CO2 = pCO2 = 0 bar
To Find:
Equilibrium partial pressures of CO and CO2
Explanation:
1. Writing the Equilibrium Expression:
The given reaction is CO2(g) + C(s) ⇌ 2CO(g)
The equilibrium constant expression for this reaction is given by Kp = (pCO)^2 / pCO2
2. Setting up the Equation:
We are given that Kp = 3.0. Substituting this value into the equilibrium constant expression, we get:
3.0 = (pCO)^2 / pCO2
3. Solving for the Equilibrium Partial Pressures:
Let's assume the equilibrium partial pressures of CO and CO2 to be x and y respectively.
According to the equation, we can write:
3.0 = x^2 / y
4. Using the Initial Partial Pressures:
At the start of the reaction, the partial pressure of CO is 0.48 bar and the partial pressure of CO2 is 0 bar.
Substituting these values into the equation, we get:
3.0 = (0.48)^2 / 0
5. Analyzing the Equation:
The equation shows that the denominator is zero, which means that the partial pressure of CO2 is zero at the start. This indicates that the reaction has not yet occurred.
6. Reaction Direction:
To determine the direction in which the reaction will proceed, we need to compare the initial partial pressures with the equilibrium partial pressures.
Since the initial partial pressure of CO2 is zero and the equilibrium constant expression contains pCO2 in the denominator, the reaction will proceed in the forward direction, converting CO2 into CO.
7. Calculating the Equilibrium Partial Pressures:
To find the equilibrium partial pressures of CO and CO2, we need to solve the equation:
3.0 = x^2 / y
8. Conclusion:
Since the reaction proceeds in the forward direction, the equilibrium partial pressure of CO2 will be zero, and the equilibrium partial pressure of CO can be calculated by solving the equation.
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The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.?
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The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.?.
The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.?.
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