Question

Equilibrium constant for the given reaction is $K={10}^{20}$ at temperature $300K$
$A\left(s\right)+2B(aq.)\rightleftharpoons 2C\left(s\right)+D(aq.)$; $K={10}^{20}$
The equilibrium concentration of B starting with a mixture of $1mol/lit$ of A and $1/3mol/lit$ of B at $300K$ is ________.

• A. $4.71\times {10}^{-11}$
• B. $2\times {10}^{-10}$
• C. $2\times {10}^{-11}$
• D. ${10}^{-10}$

Answer 1

The correct option is A $4.71\times {10}^{-11}$

$A\left(s\right)+2B\left(aq\right)\rightleftharpoons 2C\left(s\right)+D\left(aq\right)$

Initially 1 $\frac{1}{3}$ 0 0

At eq. $1-x$ $\frac{1}{3}-x$ $2x$ $x$
$\frac{2}{3}$ $\approx a$ $\frac{2}{3}$ $\frac{1}{3}$

Since, $x\approx \frac{1}{3}$ ( As $B$ gets nearly completely consumed)

Now,

$K_c=\frac{\left[C{]}^2\right[D]}{\left[A\right][B{]}^2}$

${10}^{20}=\frac{\frac{2^2\times 1}{3^2\times 3}}{\frac{2}{3}\times [B{]}^2}$ $\Rightarrow$ ${10}^{20}=\frac{\frac{2}{9}}{a^2}$ $\Rightarrow$ $a^2=\frac{2}{9\times {10}^{20}}=\frac{2\times {10}^{-20}}{9}$

$\therefore a=4.71\times {10}^{-11}$

Hence, the correct option is $\left(A\right)$