Question

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

• A. $\frac{n}{2}$
• B. $(3n-\frac{1}{2})$
• C. $(n-\frac{n}{3})$
• D. n

Answer 1

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

Given equilibrium is,
$A+B\leftrightharpoons C+D$
$\text{Initial concentration:}3nn00$
$\text{at equilibrium}3n-xn-xxx$
At equilibrium the concentration of C is equal to concentration of B
$(n-x)=xn=2xx=\frac{n}{2}$
So, the concentration of $C\left(\frac{n}{2}\right)$ is equal to the concentration of D.
Thus the concentration of $D\left(x\right)=\frac{n}{2}$