Question

In the reaction: $A+B\rightleftharpoons C+D$. The initial concentration of $A$ and $B$ are $1M$ each. The value of $K_c$ is ${10}^8$. What is the equilibrium concentration of $A$?

• A. $1\times {10}^{-4}M$
• B. $2\times {10}^{4}M$
• C. $0.005M$
• D. $0.0025M$

Answer 1

The correct option is A $1\times {10}^{-4}M$

In the reaction: $A+B\rightleftharpoons C+D$. The initial concentration of $A$ and $B$ are $1M$ each. The value of $K_c$ is ${10}^8$. The equilibrium concentration of $A$ is $2\times {10}^{-4}M$.
Let $xM$ of A react with $xM$ of B to reach equilibrium.
$xM$ C and $xM$ of D will be formed.
$(1-x)M$ of A and $(1-x)M$ of B will remain.
The equilibrium constant
$K_c=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$
${10}^8=\frac{x\times x}{(1-x)\times (1-x)}$
${10}^8=(\frac{x}{(1-x)}{)}^2$
${10}^4=\frac{x}{(1-x)}$
${10}^4-{10}^{4}x=x$
${10}^4={10}^{4}x+x$
$x=0.9999$
$[A{]}_{eq}=1-x=1-0.9999=1\times {10}^{-4}M$