One mole of ethanol is treated with one mole of ethanoic acid at 25° C. Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be a) 1 b) 2 c) 3 d) 4

Question

One mole of ethanol is treated with one mole of ethanoic acid at 25° C. Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be a) 1 b) 2 c) 3 d) 4 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/ea3d6a7e502c4012c9f29ad6e3850d8e-.png" /> K=([ CH 3 COOC 2 H 5][ H 2 O ]/[ C 2 H 5 OH ][ CH 3 COOH ])=((1/2) × (1/2)/(1)2 × (1)2=1

Q. One mole of ethanol is treated with one mole of ethanoic acid at $2 5^{\circ} C$ . Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be
a) 1
b) 2
c) 3
d) 4

$K = \frac{[ C H _{3} COO C _{2} H _{5} ] [ H _{2} O ]}{[ C _{2} H _{5} O H ] [ C H _{3} COO H ]} = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}} = 1$

Q. One mole of ethanol is treated with one mole of ethanoic acid at 25∘C. Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be a) 1 b) 2 c) 3 d) 4

K=[C2​H5​OH][CH3​COOH][CH3​COOC2​H5​][H2​O]​=21​×21​21​×21​​=1

Q. One mole of ethanol is treated with one mole of ethanoic acid at $2 5^{\circ} C$ . Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be
a) 1
b) 2
c) 3
d) 4

$K = \frac{[ C H _{3} COO C _{2} H _{5} ] [ H _{2} O ]}{[ C _{2} H _{5} O H ] [ C H _{3} COO H ]} = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}} = 1$