Question
C2H5OH(l) + C2H5CHOOH(l) ⇌ C2H5COOC2H5(l) + H2O(l)
7.7 mol of C2H5OH(l) and 7.32 mol of C2H5COOH(l) are reacted in a beaker, 4.88 mol of C2H5COOC2H5(l) are present at equilibrium and the total volume of liquid is exactly 2.00 L. What the K eq for this system?
7.7 mol of C2H5OH(l) and 7.32 mol of C2H5COOH(l) are reacted in a beaker, 4.88 mol of C2H5COOC2H5(l) are present at equilibrium and the total volume of liquid is exactly 2.00 L. What the K eq for this system?
Answers
GPT 3.5
To calculate the K_eq for this system, we need to set up the equation using the equilibrium concentrations.
The balanced equation for the reaction is:
C2H5OH(l) + C2H5COOH(l) ⇌ C2H5COOC2H5(l) + H2O(l)
Let x be the equilibrium concentration of C2H5COOC2H5(l).
The initial concentration of C2H5OH(l) is 7.7 mol / 2.00 L = 3.85 M.
The initial concentration of C2H5COOH(l) is 7.32 mol / 2.00 L = 3.66 M.
The equilibrium concentration of C2H5OH(l) is 3.85 M - x.
The equilibrium concentration of C2H5COOH(l) is 3.66 M - x.
The equilibrium concentration of C2H5COOC2H5(l) is x.
The equilibrium concentration of H2O(l) is 0 M because it is a liquid and does not affect the equilibrium expression.
The K_eq expression for this reaction is:
K_eq = [C2H5COOC2H5] / ([C2H5OH] * [C2H5COOH])
K_eq = x / (3.85 - x) * (3.66 - x)
Given that the equilibrium concentration of C2H5COOC2H5(l) is 4.88 M, we can substitute this value into the equation:
4.88 = x / (3.85 - x) * (3.66 - x)
Now we can solve for x:
4.88 * (3.85 - x) * (3.66 - x) = x
Simplifying the equation:
4.88 * (3.85 - x) * (3.66 - x) = x
14.2512 - 4.88x - 14.2312x + 4.88x^2 = x
4.88x^2 - 18.1112x + 14.2512 = x
4.88x^2 - 19.9912x + 14.2512 = 0
Using the quadratic formula, we can solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
a = 4.88, b = -19.9912, c = 14.2512
x = (-(-19.9912) ± √((-19.9912)^2 - 4*4.88*14.2512)) / (2*4.88)
x = (19.9912 ± √(399.6484 - 280.69952)) / 9.76
x = (19.9912 ± √118.94888) / 9.76
Calculating both roots:
x ≈ 2.20 M or x ≈ 0.002 M
Since x must be less than the initial concentration of C2H5COOH, we can discard the root x ≈ 2.20 M.
Therefore, the equilibrium concentration of C2H5COOC2H5(l) is approximately 0.002 M.
Now we can calculate the K_eq using this value:
K_eq = 0.002 / (3.85 - 0.002) * (3.66 - 0.002)
K_eq ≈ 0.002 / 3.848 * 3.658
K_eq = 1.6951 * 10^-4
Therefore, the K_eq for this system is approximately 1.6951 * 10^-4.
The balanced equation for the reaction is:
C2H5OH(l) + C2H5COOH(l) ⇌ C2H5COOC2H5(l) + H2O(l)
Let x be the equilibrium concentration of C2H5COOC2H5(l).
The initial concentration of C2H5OH(l) is 7.7 mol / 2.00 L = 3.85 M.
The initial concentration of C2H5COOH(l) is 7.32 mol / 2.00 L = 3.66 M.
The equilibrium concentration of C2H5OH(l) is 3.85 M - x.
The equilibrium concentration of C2H5COOH(l) is 3.66 M - x.
The equilibrium concentration of C2H5COOC2H5(l) is x.
The equilibrium concentration of H2O(l) is 0 M because it is a liquid and does not affect the equilibrium expression.
The K_eq expression for this reaction is:
K_eq = [C2H5COOC2H5] / ([C2H5OH] * [C2H5COOH])
K_eq = x / (3.85 - x) * (3.66 - x)
Given that the equilibrium concentration of C2H5COOC2H5(l) is 4.88 M, we can substitute this value into the equation:
4.88 = x / (3.85 - x) * (3.66 - x)
Now we can solve for x:
4.88 * (3.85 - x) * (3.66 - x) = x
Simplifying the equation:
4.88 * (3.85 - x) * (3.66 - x) = x
14.2512 - 4.88x - 14.2312x + 4.88x^2 = x
4.88x^2 - 18.1112x + 14.2512 = x
4.88x^2 - 19.9912x + 14.2512 = 0
Using the quadratic formula, we can solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
a = 4.88, b = -19.9912, c = 14.2512
x = (-(-19.9912) ± √((-19.9912)^2 - 4*4.88*14.2512)) / (2*4.88)
x = (19.9912 ± √(399.6484 - 280.69952)) / 9.76
x = (19.9912 ± √118.94888) / 9.76
Calculating both roots:
x ≈ 2.20 M or x ≈ 0.002 M
Since x must be less than the initial concentration of C2H5COOH, we can discard the root x ≈ 2.20 M.
Therefore, the equilibrium concentration of C2H5COOC2H5(l) is approximately 0.002 M.
Now we can calculate the K_eq using this value:
K_eq = 0.002 / (3.85 - 0.002) * (3.66 - 0.002)
K_eq ≈ 0.002 / 3.848 * 3.658
K_eq = 1.6951 * 10^-4
Therefore, the K_eq for this system is approximately 1.6951 * 10^-4.