Question

For the reaction $A\left(g\right)+2B\left(g\right)\to C\left(g\right)+D\left(g\right);K_c={10}^{12}$.
If the initial moles of $A,B,C$ and $D$ are $0.5,1,0.5$ and $3.5$ moles respectively in a one litre vessel. What is the equilibrium concentration of $B$?

• A. ${10}^{-4}$
• B. $2\times {10}^{-4}$
• C. $4\times {10}^{-4}$
• D. $8\times {10}^{-4}$

Answer 1

The correct option is B $2\times {10}^{-4}$

$A\left(g\right)+2B\left(g\right)\to C\left(g\right)+D\left(g\right)$ $K_C={10}^{12}$

At $0.5$ $1$ $0.5$ $0.5$

$t=0$

At $0.5-\alpha$ $1-2\alpha$ $0.5+\alpha$ $3.5+\alpha$

$t=t_{eq}$

Since $K_C={10}^{12}$ very large $\alpha \approx 0.5$

Let $0.5-\alpha$ be $\left(x\right)$

So $\left[A\right]=x$

$\left[B\right]=x$

$\left[C\right]\approx 1$

$\left[D\right]\approx 4$

$K_C=\frac{\left[C\right]\left[D\right]}{{\left[B\right]}^2\left[A\right]}\frac{1\times 4}{{4x}^2\times x}={10}^{+12}$

$x={10}^{-4}$

$\left[B\right]=2x=2\times {10}^{-4}M$