In the system $A (s) ⇌ 2 B (g) + 3 C (g)$ , the concentration of $C$ at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of $B$ to change to:

two times the original value

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one half of its original value

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2 \sqrt{2}

times its original value

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\frac{1}{2 \sqrt{2}}

times its original value

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Solution

The correct option is D $\frac{1}{2 \sqrt{2}}$ times its original value
As we know,
$K_{c} = {[C]}^{3} {[B]}^{2}$ if $[C] = [2 C]$ , then to have $K_{c}$ constant $[B]$ should be reduced to $\frac{[B]}{2 \sqrt{2}}$

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In the system A(s)⇌2B(g)+3C(g), the concentration of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to change to:

The correct option is D 12√2 times its original valueAs we know,Kc=[C]3[B]2 if [C]=[2C], then to have Kc constant [B] should be reduced to [B]2√2

In the system $A (s) ⇌ 2 B (g) + 3 C (g)$ , the concentration of $C$ at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of $B$ to change to:

The correct option is D $\frac{1}{2 \sqrt{2}}$ times its original value
As we know,
$K_{c} = {[C]}^{3} {[B]}^{2}$ if $[C] = [2 C]$ , then to have $K_{c}$ constant $[B]$ should be reduced to $\frac{[B]}{2 \sqrt{2}}$