To find the equilibrium concentration of C2H5OH, we need to use the information given about the initial moles of C2H4 and H2O and the value of Kc.
The balanced chemical equation for the reaction is:
C2H4 + H2O ⇌ C2H5OH
Let's use the following notation:
[C2H4] = concentration of C2H4 (in mol/L)
[H2O] = concentration of H2O (in mol/L)
[C2H5OH] = concentration of C2H5OH (in mol/L)
Given:
Initial moles of C2H4 = 115 mol
Initial moles of H2O = 110 mol
Kc = 300.00
Since the reaction vessel has a volume of 5000L, we can calculate the initial concentrations:
[C2H4] = moles of C2H4 / volume of reaction vessel = 115 mol / 5000 L = 0.023 mol/L
[H2O] = moles of H2O / volume of reaction vessel = 110 mol / 5000 L = 0.022 mol/L
Using the equation for Kc:
Kc = [C2H5OH] / ([C2H4] * [H2O])
Let's substitute the values into the equation:
300.00 = [C2H5OH] / (0.023 * 0.022)
Simplifying the equation:
300.00 = [C2H5OH] / 0.000506
[C2H5OH] = 300.00 * 0.000506
Calculating the value of [C2H5OH]:
[C2H5OH] = 0.1518 mol/L
Therefore, the equilibrium concentration of C2H5OH in the reaction vessel is 0.1518 mol/L.
A 5000L reaction vessel contains 115 mol of C2H4 and 110 mol of H2O, with the Kc of 300.00. What is the equilibrium concentration of C2H5OH ?
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