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 2, it will cause the equilibrium concentration of $B$ to change to:

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

Answer 1

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

The equilibrium reaction is as shown below:
$A_{\left(s\right)}\rightleftharpoons 2B_{\left(g\right)}+3C_{\left(g\right)}$
$1-x2x3x$
Th expression for the equilibrium constant is $K_c=4x^2\times 27x^3$.
Now the concentration of C is increased from $3xto6x$.
The expression for the equilibrium constant becomes $108x^5=4x^2\times c^2\times 216x^3$.
Thus $108=c^2\times 216\times 4$ or $c^2=\frac{108}{4\times 216}$.
Hence, $c=\frac{1}{2\sqrt{2}}$.