Question

Consider the general hypothetical reaction:
$A\left(s\right)\leftrightharpoons 2B\left(g\right)+3C\left(g\right)$
If the concentration of C at equilibrium is doubled, then after the equilibrium is re-established, the concentration of B will be :

• A. Twice of its original value
• B. Half of its original value
• C. $2\sqrt{2}$ times of original value
• D. $\frac{1}{2\surd 2}$ times the original value

Answer 1

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

The given reaction is,
$A\left(s\right)\leftrightharpoons 2B\left(g\right)+3C\left(g\right)$
So the equilibrium constant will be,

$K_c=\left[B{]}^2\right[C{]}^3$....(1)

Now, after doubling the concentration of C, let new concentrations are $[C{]}^{/}$ and $[B{]}^{/}$.
Given, $[C{]}^{/}=2[C]$
So,
$K_c={\left(\right[B{]}^{/})}^2{\left(2\right[C\left]\right)}^3$
$\Rightarrow K_c={\left(\right[B{]}^{/})}^{2}8[C{]}^3$ ...(2)

From (1) and (2),
$\left[B{]}^2\right[C{]}^3={\left(\right[B{]}^{/})}^{2}8[C{]}^3$
$\Rightarrow [B{]}^{/}=\frac{1}{2\sqrt{2}}[B]$