A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL of 0.200M (mol/L) compound B. At equilibrium, the concentration of C is 0.0454M. Calculate the equilibrium constant, K, for this reaction.A (aq) + 2B (aq) <---> C (aq)The initial concentrations of A and B are not 0.150 M and 0.200 M
Question
A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL of 0.200M (mol/L) compound B. At equilibrium, the concentration of C is 0.0454M. Calculate the equilibrium constant, K, for this reaction.A (aq) + 2B (aq) <---> C (aq)The initial concentrations of A and B are not 0.150 M and 0.200 M
Solution
To calculate the equilibrium constant, K, for this reaction, we need to use the concentrations of the reactants and products at equilibrium.
Given:
- Volume of compound A (A) = 30.0 mL
- Concentration of compound A (A) = 0.150 M
- Volume of compound B (B) = 25.0 mL
- Concentration of compound B (B) = 0.200 M
- Concentration of compound C (C) at equilibrium = 0.0454 M
The balanced equation for the reaction is: A (aq) + 2B (aq) ⇌ C (aq)
To calculate the equilibrium constant, K, we need to use the formula: K = [C]^c / ([A]^a * [B]^b)
Where [C], [A], and [B] represent the concentrations of C, A, and B at equilibrium, respectively, and a, b, and c are the stoichiometric coefficients of the balanced equation.
In this case, the stoichiometric coefficients are: a = 1 (for A) b = 2 (for B) c = 1 (for C)
Plugging in the values, we have: K = (0.0454) / ((0.150)^1 * (0.200)^2)
Calculating this expression, we find the value of K.
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