Write the equilibrium expression, Kc, for the following chemical reaction. 2C6Halg) + 1502(g) * 12CO2(g) + 6H2O(g). [co,] [H,0] А. ['o] [°n°ɔ] |co.]" [1,0] [co,] [1,0] [co,]" [1,0] D. B.
Write the equilibrium expression, Kc, for the following chemical reaction. 2C6Halg) + 1502(g) * 12CO2(g) + 6H2O(g). [co,] [H,0] А. ['o] [°n°ɔ] |co.]" [1,0] [co,] [1,0] [co,]" [1,0] D. B.
Chemistry
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Equilibrium Expression for a Chemical Reaction**
For the chemical reaction:
\[ 2\text{C}_6\text{H}_6(g) + 15\text{O}_2(g) \rightleftharpoons 12\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \]
We are asked to identify the correct equilibrium expression, \( K_c \).
**Options:**
A. \(\frac{[\text{CO}_2][\text{H}_2\text{O}]}{[\text{C}_6\text{H}_6][\text{O}_2]}\)
B. \(\frac{[\text{CO}_2]^{12} [\text{H}_2\text{O}]^{6}}{[\text{C}_6\text{H}_6]^{2} [\text{O}_2]^{15}}\)
C. \(\frac{[\text{C}_6\text{H}_6][\text{O}_2]}{[\text{CO}_2][\text{H}_2\text{O}]}\)
D. \(\frac{[\text{C}_6\text{H}_6]^{2} [\text{O}_2]^{15}}{[\text{CO}_2]^{12} [\text{H}_2\text{O}]^{6}}\)
**Explanation:**
The equilibrium constant expression (\( K_c \)) for a reversible reaction is given by the concentrations of the products raised to the power of their coefficients, divided by the concentrations of the reactants, also raised to the power of their coefficients. Therefore, the correct expression for this reaction is:
\[ K_c = \frac{[\text{CO}_2]^{12} [\text{H}_2\text{O}]^{6}}{[\text{C}_6\text{H}_6]^{2} [\text{O}_2]^{15}} \]
Thus, the correct answer is **Option B**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2a163f18-29bc-4cae-aa7e-1a32645d5b83%2F9d1bd056-17e2-41ae-a158-064d529d5e01%2Fxe3mqyc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Expression for a Chemical Reaction**
For the chemical reaction:
\[ 2\text{C}_6\text{H}_6(g) + 15\text{O}_2(g) \rightleftharpoons 12\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \]
We are asked to identify the correct equilibrium expression, \( K_c \).
**Options:**
A. \(\frac{[\text{CO}_2][\text{H}_2\text{O}]}{[\text{C}_6\text{H}_6][\text{O}_2]}\)
B. \(\frac{[\text{CO}_2]^{12} [\text{H}_2\text{O}]^{6}}{[\text{C}_6\text{H}_6]^{2} [\text{O}_2]^{15}}\)
C. \(\frac{[\text{C}_6\text{H}_6][\text{O}_2]}{[\text{CO}_2][\text{H}_2\text{O}]}\)
D. \(\frac{[\text{C}_6\text{H}_6]^{2} [\text{O}_2]^{15}}{[\text{CO}_2]^{12} [\text{H}_2\text{O}]^{6}}\)
**Explanation:**
The equilibrium constant expression (\( K_c \)) for a reversible reaction is given by the concentrations of the products raised to the power of their coefficients, divided by the concentrations of the reactants, also raised to the power of their coefficients. Therefore, the correct expression for this reaction is:
\[ K_c = \frac{[\text{CO}_2]^{12} [\text{H}_2\text{O}]^{6}}{[\text{C}_6\text{H}_6]^{2} [\text{O}_2]^{15}} \]
Thus, the correct answer is **Option B**.
Expert Solution
Introduction
The reaction is said to be in equilibrium when the rate of forward reaction is same as rate of backward reaction.
we can say that amount of product is formed is equal to the amount of product breakdown.
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