At a certain temperature, the equilibrium constant for the chemical reaction shown is 5.26 x 10-3. At equilibrium, the concentration of AB is 2.825 M, the concentration of BC is 1.425 M, and the concentration of AC is 0.160 M. Calculate the concentration of B at equilibrium. AB(aq) + BC(aq) = AC(aq) + 2 B(aq) [B] = M
At a certain temperature, the equilibrium constant for the chemical reaction shown is 5.26 x 10-3. At equilibrium, the concentration of AB is 2.825 M, the concentration of BC is 1.425 M, and the concentration of AC is 0.160 M. Calculate the concentration of B at equilibrium. AB(aq) + BC(aq) = AC(aq) + 2 B(aq) [B] = M
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Chapter1: Chemical Foundations
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Question
![### Chemical Equilibrium Problem
At a certain temperature, the equilibrium constant (\(K_c\)) for the chemical reaction shown below is \(5.26 \times 10^{-3}\). At equilibrium, the concentration of reactant AB is 2.825 M, the concentration of reactant BC is 1.425 M, and the concentration of product AC is 0.160 M. Calculate the concentration of product B at equilibrium.
The balanced chemical equation for the reaction is:
\[ \text{AB}(aq) + \text{BC}(aq) \rightleftharpoons \text{AC}(aq) + 2\text{B}(aq) \]
**Given:**
- \(K_c = 5.26 \times 10^{-3}\)
- \([\text{AB}] = 2.825 \, \text{M}\)
- \([\text{BC}] = 1.425 \, \text{M}\)
- \([\text{AC}] = 0.160 \, \text{M}\)
**To find:**
- \([\text{B}] \) at equilibrium.
**Equation for Equilibrium Constant:**
\[ K_c = \frac{[\text{AC}][\text{B}]^2}{[\text{AB}][\text{BC}]} \]
**Substitute the known concentrations and equilibrium constant:**
\[ 5.26 \times 10^{-3} = \frac{(0.160)[\text{B}]^2}{(2.825)(1.425)} \]
**Solve for \([\text{B}]\):**
Note: Solving the above equation requires algebraic manipulation and potentially the use of a quadratic formula depending on the complexity of the equation resulting from the substitution of the known values.
Fill in the final calculated concentration:
\[ [\text{B}] = \_\_\_\_\_ \text{ M} \]
Interact with the educational content, solve on your own or follow guided steps to reach the final answer.
This mathematical approach reinforces understanding of equilibrium concepts and the application of the equilibrium constant in solving for unknown concentrations in chemical reactions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c7c097a-a119-48b6-b238-faf2426c9b82%2Fea84fe00-6ea4-4840-950b-0cd39027b146%2F3bmkrfl_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemical Equilibrium Problem
At a certain temperature, the equilibrium constant (\(K_c\)) for the chemical reaction shown below is \(5.26 \times 10^{-3}\). At equilibrium, the concentration of reactant AB is 2.825 M, the concentration of reactant BC is 1.425 M, and the concentration of product AC is 0.160 M. Calculate the concentration of product B at equilibrium.
The balanced chemical equation for the reaction is:
\[ \text{AB}(aq) + \text{BC}(aq) \rightleftharpoons \text{AC}(aq) + 2\text{B}(aq) \]
**Given:**
- \(K_c = 5.26 \times 10^{-3}\)
- \([\text{AB}] = 2.825 \, \text{M}\)
- \([\text{BC}] = 1.425 \, \text{M}\)
- \([\text{AC}] = 0.160 \, \text{M}\)
**To find:**
- \([\text{B}] \) at equilibrium.
**Equation for Equilibrium Constant:**
\[ K_c = \frac{[\text{AC}][\text{B}]^2}{[\text{AB}][\text{BC}]} \]
**Substitute the known concentrations and equilibrium constant:**
\[ 5.26 \times 10^{-3} = \frac{(0.160)[\text{B}]^2}{(2.825)(1.425)} \]
**Solve for \([\text{B}]\):**
Note: Solving the above equation requires algebraic manipulation and potentially the use of a quadratic formula depending on the complexity of the equation resulting from the substitution of the known values.
Fill in the final calculated concentration:
\[ [\text{B}] = \_\_\_\_\_ \text{ M} \]
Interact with the educational content, solve on your own or follow guided steps to reach the final answer.
This mathematical approach reinforces understanding of equilibrium concepts and the application of the equilibrium constant in solving for unknown concentrations in chemical reactions.
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