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When 2A (g) <-----> B(g) + C(g) [exothermic] is at equilibrium the concentration

ID: 952219 • Letter: W

Question

When 2A (g) <-----> B(g) + C(g) [exothermic] is at equilibrium the concentration of B

a. increase

b. decreases

c. is unchanged

d. changes randomly

e. cannot predict

Concentration of C is increased, concentration of A

a. increase

b. decreases

c. is unchanged

d. changes randomly

e. cannot predict

Concentration of C is increased, concentration of B

a. increase

b. decreases

c. is unchanged

d. changes randomly

e. cannot predict

If temp is raised the concentration of C

a. increase

b. decreases

c. is unchanged

d. changes randomly

e. cannot predict

If volume of reaction is decreased concentration of A will

a. increase

b. decreases

c. is unchanged

d. changes randomly

e. cannot predict

when reaction is at equilibrium

a. [a]=[b]=[c]

b. [a],[b]&[c] depend on Kc

c. Kc < Qc

d. Kc > Qc

e. cannot predict

Please briefly explain each

Explanation / Answer

ANSWER

2A(g) <-----> B(g) + C(g)

The questions are based on Le Chatliers Principle, which states that if the equilibrium is disturbed the equilibrium changes in the direction that can offset the disturbance.

(A) Atequilibrium the concentration of the reactants and products dies not change.

Hence option (c) is correct answer

(B) Concentration of C is increased, concentration of A

If concentration of C is increased the equilibrium will move in backward direction to offset the change. Hence concentration of A will increase. Option (a) is correct answer.

(C) Concentration of C is increased, concentration of B

Option (b) is correct answer. Because the reaction will move in the backward reaction.

(D) If temp is raised the concentration of C.

Forward reaction is exothermic hance backward reaction must be endothermic. When temperature is increased backward reaction will be favoured so as absorb the extra heat. Hence concentration of C decreases. Option (b) is correct answer.

(E) There will be effect of volume change on the equilibrium, because the no. of moles of reactants (2) is equal to the number of moles of the products (2).

(F) Option (b) is correct answer.

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