Question

The reaction $A+B\rightleftharpoons C+D$ is studied in a one litre vessel at ${250}^{0}C$. The initial concentration of $A$ was 3n and of $B$ was n. After equilibrium was attained then equilibrium concentration of $B$. What is the concentration of $D$ at equilibrium : -

• A. $\frac{n}{2}$
• B. $(3n-\frac{n}{2})$
• C. $(n+\frac{n}{2})$
• D. n

Answer 1

The correct option is A $\frac{n}{2}$

We have the given reaction.

$A+B\leftrightarrows C+D$

$att=03nn00$

$att=t3n-xn-xxx$

at equilibrium moles of $B$ is equal to moles of $C$

$\therefore n-x=x$

$x=\frac{n}{2}$

$A+B\leftrightarrows C+D$

$3n-\frac{n}{2}n-\frac{n}{2}\frac{n}{2}\frac{n}{2}$

Hence Number of Moles of $D=\frac{n}{2}$

So, the option $\left(A\right)$ is the correct answer.