Question

Guys pls help me in this question Posting irrelevant answers Pls stop it​

Answer 1

Answer:

$\sf\underline{question}$

The equilibrium constant of the reaction, CH_3COOH + C_2H_5OH $\sf\rightleftharpoons$ CH_3COOC_2H_5 + H_2O is 4. If one mole of each acetic and ethyl alcohol are heated in presence of a little concentrated H_2SO_4, at equilibrium the amount of ester present is :

$\huge\sf\underline{solution}$

according to ICE table :-

note :- refer to above attachment.

k_c = 4

$\sf\dfrac{x×x}{(1-x)(1-x)}$ = 4

$\sf\dfrac{x²}{(1-x)²}$ = 4

x² = 4×(1-x)²

x² = 4[1-2x+x²]

x² = 4-8x+4x²

4x²-8x+4-x² = 0

3x²-8x+4 = 0

3x²-2x-6x+4 = 0

x(3x-2)-2(3x-2) = 0

x = 2, x = 2/3

if we take x = 2 then equilibrium will be negative in ICE so we shouldn't take that.

since, x = 2/3

Answer 2

Answer:

according to ICE table :-

note :- refer to above attachment.

k_c = 4

\sf\dfrac{x×x}{(1-x)(1-x)}

(1−x)(1−x)

x×x

= 4

\sf\dfrac{x²}{(1-x)²}

(1−x)²

x²

= 4

x² = 4×(1-x)²

x² = 4[1-2x+x²]

x² = 4-8x+4x²

4x²-8x+4-x² = 0

3x²-8x+4 = 0

3x²-2x-6x+4 = 0

x(3x-2)-2(3x-2) = 0

x = 2, x = 2/3

if we take x = 2 then equilibrium will be negative in ICE so we shouldn't take that.

since, x = 2/3