[4] At 25 °C, the equilibrium constant for the reaction: 2CO(g) +O2(g) = 2CO2(g) Kc = 5.0x1018. What is the value of K. at this temperature for the following reactions: a) 4CO(g) +202(g) = 4CO2(g) b) 2CO₂(g) = 2CO(g) + O2(g) c) CO₂(g) = CO(g) +1/2 O₂(g)
[4] At 25 °C, the equilibrium constant for the reaction: 2CO(g) +O2(g) = 2CO2(g) Kc = 5.0x1018. What is the value of K. at this temperature for the following reactions: a) 4CO(g) +202(g) = 4CO2(g) b) 2CO₂(g) = 2CO(g) + O2(g) c) CO₂(g) = CO(g) +1/2 O₂(g)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Equilibrium Constants in Chemical Reactions**
At 25°C, the equilibrium constant \( K_c \) for the reaction \( 2\text{CO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{CO}_2(g) \) is \( K_c = 5.0 \times 10^{18} \).
We need to determine the value of \( K_c \) at this temperature for the following reactions:
a) \( 4\text{CO}(g) + 2\text{O}_2(g) \rightleftharpoons 4\text{CO}_2(g) \)
b) \( 2\text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) + \text{O}_2(g) \)
c) \( \text{CO}_2(g) \rightleftharpoons \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \)
---
**Understanding Reactions and Equilibrium Constants**
The equilibrium constant, \( K_c \), for a reaction at a given temperature is a measure of the ratio of the concentrations of products to reactants at equilibrium. For a general reaction \( \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD} \), the equilibrium constant expression is given by:
\[ K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \]
When manipulating reactions, keep in mind:
- Reversing a reaction inverts the \( K_c \) value
- Multiplying the coefficients of a reaction by a factor \( n \) raises the \( K_c \) to the power \( n \)
Apply these rules to adjust for the changes in reactions a, b, and c as needed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc20a2323-0224-4452-b8ac-6962646b54a1%2F4e590f11-84a5-4b81-901f-8e7e8224d548%2F4exhgy5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constants in Chemical Reactions**
At 25°C, the equilibrium constant \( K_c \) for the reaction \( 2\text{CO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{CO}_2(g) \) is \( K_c = 5.0 \times 10^{18} \).
We need to determine the value of \( K_c \) at this temperature for the following reactions:
a) \( 4\text{CO}(g) + 2\text{O}_2(g) \rightleftharpoons 4\text{CO}_2(g) \)
b) \( 2\text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) + \text{O}_2(g) \)
c) \( \text{CO}_2(g) \rightleftharpoons \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \)
---
**Understanding Reactions and Equilibrium Constants**
The equilibrium constant, \( K_c \), for a reaction at a given temperature is a measure of the ratio of the concentrations of products to reactants at equilibrium. For a general reaction \( \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD} \), the equilibrium constant expression is given by:
\[ K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \]
When manipulating reactions, keep in mind:
- Reversing a reaction inverts the \( K_c \) value
- Multiplying the coefficients of a reaction by a factor \( n \) raises the \( K_c \) to the power \( n \)
Apply these rules to adjust for the changes in reactions a, b, and c as needed.
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