Sure! Let's go through the steps to calculate the equilibrium constant [tex]\( K_c \)[/tex] for the given chemical reaction in a detailed manner.
Given data:
- [tex]\( [\text{CH}_3\text{COOC}_2\text{H}_5] = \frac{2}{3} \)[/tex]
- [tex]\( [\text{H}_2\text{O}] = \frac{2}{3} \)[/tex]
- [tex]\( [\text{CH}_3\text{COOH}] = \frac{1}{3} \)[/tex]
- [tex]\( [\text{C}_2\text{H}_5\text{OH}] = \frac{1}{3} \)[/tex]
The equilibrium constant [tex]\( K_c \)[/tex] for a reaction is given by the equation:
[tex]\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \][/tex]
Substituting in the concentrations:
[tex]\[ K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]} \][/tex]
We need to perform the multiplications and the division step-by-step:
1. Calculate the numerator:
[tex]\[ \left[\text{CH}_3\text{COOC}_2\text{H}_5\right] \times \left[\text{H}_2\text{O}\right] = \left(\frac{2}{3}\right) \times \left(\frac{2}{3}\right) = \frac{4}{9} \][/tex]
2. Calculate the denominator:
[tex]\[ \left[\text{CH}_3\text{COOH}\right] \times \left[\text{C}_2\text{H}_5\text{OH}\right] = \left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right) = \frac{1}{9} \][/tex]
3. Divide the numerator by the denominator to obtain [tex]\( K_c \)[/tex]:
[tex]\[ K_c = \frac{\frac{4}{9}}{\frac{1}{9}} = \frac{4}{9} \times \frac{9}{1} = 4 \][/tex]
Therefore, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction is:
[tex]\[ K_c = 4 \][/tex]
So, in summary, we've found the equilibrium constant step-by-step from the given concentrations of the reactants and products.
