Question

In: Chemistry

For the reaction shown below, the initial concentrations for reactants and products are as follows: [A]...

For the reaction shown below, the initial concentrations for reactants and products are as follows: [A] = 0.27 M, [B] = 0.0640 M, [C] = 0.0134 M, and [D] = 1.47 M. If the equilibrium constant for the reaction is K_C = 0.802, which direction does the reaction have to shift to reach equilibrium?

A + 3B ⇌ 2C + 3D

Enter 1 if it shifts left, 2 if it shifts right, or 3 if it's already at equilibrium.

Solutions

Expert Solution

Answer:-

Given:-

molar concentration of A i.e [A] = 0.27 M

molar concentration of B i.e [B] = 0.0640 M

molar concentration of C i.e [C] = 0.0134 M

molar concentration of D i.e [D] = 1.47 M

equilibrium constant (Kc) = 0.802

As we know that

A + 3B 2C + 3D

therefore

equilibrium constant (Kc) = [C]2[D]3 / [A][B]3

equilibrium constant (Kc) = (0.0134)2 (1.47)3 / (0.27)(0.0640)3

equilibrium constant (Kc) = 0.00017956‬ 3.176523‬ / 0.27 0.000262144‬

equilibrium constant (Kc) = 0.00057037646988‬‬ / 0.00007077888‬

equilibrium constant (Kc) = 8.05

As mentioned in above value of equilibrium constant (Kc) i.e calculated value of equilibrium constant (Kc) is equivalent to the given value of equilibrium constant (Kc).

calculated value of equilibrium constant (Kc) given value of equilibrium constant (Kc)

8.05 0.802

Therefore we can say that the reaction is already at equilibrium. Since we know that the value of equilibrium constant (Kc) indicates that following conclusions.

(I) - Greater value of equilibrium constant (Kc) indicates that the reaction is proceeded to a greater extent in the forward direction i.e formation of product (C and D).

(II) - Lower value of equilibrium constant (Kc) indicates that the reaction is proceeded to a greater extent in the backward direction i.e formation of reactants (A and B).

(III) - If the value of equilibrium constant (Kc) of the reaction is equal to 1 or given value of equilibrium constant (Kc) then it means that reaction is at equilibrium. At equilibrium, rate of forward reaction i.e formation of products (C and D) is equal to the rate of backward reaction i.e formation of reactants (A and B). In other words at equilibrium there is no formation of products or reactants.

therefore

correct answer is as follows:-

The reaction is already at equilibrium i.e 3. ( answer)

queen_honey_blossom answered 7 months ago

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