Question

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

• A. Two times the original value
• B. One half of its original value
• C. $2\sqrt{2}$ times the original value
• D. $\frac{1}{2\sqrt{2}}$ times the original value

Answer 1

The correct option is D $\frac{1}{2\sqrt{2}}$ times the original value

$Explanation:$

Given

Equation $A_{\left(s\right)}\rightleftharpoons 2B_{\left(g\right)}+3C_{\left(g\right)}$

Initial concentration $(1-x)$ $2x$ $3x$

Equilibrium constant$=K_{eq}={\left[B\right]}^2\times {\left[C\right]}^3$

$4x^2\times 27x^3=108x^5$

Given that the concentration of C Increases by a factor of 2

Then

$A_{\left(s\right)}\rightleftharpoons 2B_{\left(g\right)}+3C_{\left(g\right)}$

new concentration $(1-x)$ $2xc$ $6xc$

Equilibrium constant after change in concentration $=K_{eq}^{\prime }={\left[B^{\prime }\right]}^2\times {\left[C^{\prime }\right]}^3$

we know that equilibrium will be same through the concentration

$\Rightarrow K_{eq}=K^{\prime }eq$

$108x=4x^2\times c^2\times 27x^3$

Then

$C_2=\frac{108}{4\times 216}=\frac{1}{2\sqrt{2}}$

Hence the correct answer is option$D$.