The equilibrium constant of the reaction CH3COOH (l)+C2H5OH (l)=CH3COOC2H5+H2O is 4.if one mole of each of acetic acid and ethyl alcohol are heated in presence of little concentrated H2SO4 at equilibrium the amount of ester present is?

See above.

2/3 mole

To determine the amount of ester present at equilibrium, we need to use the equilibrium constant and the initial moles of acetic acid and ethyl alcohol.

Given:
Equilibrium constant (K) = 4
Moles of acetic acid (CH3COOH) = 1 mole
Moles of ethyl alcohol (C2H5OH) = 1 mole

The balanced equation for the reaction is:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

The equilibrium constant expression can be written as:
K = ([CH3COOC2H5][H2O]) / ([CH3COOH][C2H5OH])

Since the concentrations of water and the catalyst (H2SO4) are not included in the equilibrium constant expression, we can ignore them for this calculation.

Let's designate x as the amount of ester (CH3COOC2H5) formed. Since the stoichiometric ratio of ester to acetic acid and ethyl alcohol in the balanced equation is 1:1, the amount of acetic acid and ethyl alcohol that reacts is also x.

Using the equilibrium constant expression, we have:
4 = (x)(x) / (1 - x)(1 - x)

Simplifying the equation:
4 = x^2 / (1 - x)^2

Cross-multiplying:
4(1 - x)^2 = x^2

Expanding and simplifying:
4(1 - 2x + x^2) = x^2

4 - 8x + 4x^2 = x^2

Rearranging the equation:
3x^2 - 8x + 4 = 0

To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2-4ac)) / (2a)

In our equation, a = 3, b = -8, and c = 4.

Solving for x:
x = (-(-8) ± √((-8)^2 - 4(3)(4))) / (2(3))
x = (8 ± √(64 - 48)) / 6
x = (8 ± √16) / 6

Taking the positive value:
x = (8 + 4) / 6
x = 12 / 6
x = 2

Therefore, at equilibrium, the amount of ester (CH3COOC2H5) present is 2 moles.