Question

The reaction, $A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$ is studied in a vessel of one litre at ${250}^{o}C$. The initial concentration of $A$ was $3x$ that of $B$ was $x$. At equilibrium, concentration of $C$ was found to be equal to the equilibrium concentration of $B$. What is the concentration of $D$ at equilibrium?

• A. $\frac{x}{2}$
• B. $3x-\frac{1}{2}$
• C. $x-\frac{x}{2}$
• D. $x$

Answer 1

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

Explanation:

Given Rection is :

$A_{\left(g\right)}+B_{\left(g\right)}\rightleftharpoons C_{\left(g\right)}+D_{\left(g\right)}$

$3x$ $x$ $0$ $0$ at time $t=0$

$3x-n$ $x-n$ $n$ $n$ at time $t=t$

Moles of $B$ at equilibrium $=$ Moles of $C$

$\therefore x-n=n$

$x=2n$

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

Cocentration of each element at equilibrium is :

$\Rightarrow$ ${\left[A\right]}_{\left(g\right)}=3x-n=(3x-\frac{x}{2})=\frac{6x-x}{2}=\frac{5x}{2}$

$\Rightarrow$ ${\left[B\right]}_{\left(g\right)}=x-n=x-\frac{x}{2}$

$\Rightarrow$ ${\left[C\right]}_{\left(g\right)}=n=\frac{x}{2}$

$\Rightarrow$ ${\left[D\right]}_{\left(g\right)}=n=\frac{x}{2}$

$\therefore$ At equilibrium concentration of $D=\frac{x}{2}$

Hence the correct answer is option $A.$