An equilibrium mixture contains 2.0 moles of A(g) and 4.0 moles of B(g). Now, 2.0 moles of A(g) is added in this equilibrium mixture and the system is allowed to re-achieve equilibrium at constant volume and temperature. Now, the volume of system is doubled at constant temperature. What should be the moles of B(g) at new equilibrium? The reaction involved is

$A\left(g\right)\stackrel{}{?}2B\left(g\right)$ .

$A\left(g\right)?2B\left(g\right).$

$E_m\frac{2mol}{V\mathcal{l}}\frac{4mol}{V\mathcal{l}}$

$New{E_q}^m\frac{4-x}{V}\frac{4+2x}{V}$

$V*2\frac{4-x}{2V}\frac{4-2x}{2V}$

New Eqm $\frac{4-x-y}{2V}\frac{4+2x+2y}{2V}$

= $\frac{4-Z}{2V}=\frac{4+2Z}{2V}\left(Z=x+y\right)$

Now, KC = $\frac{{}^{}}{}$ $\frac{{\left[B\right]}^2}{\left[A\right]}=$ constant

$or,\frac{{\left(\frac{4}{V}\right)}^2}{\left(\frac{2}{V}\right)}=\frac{\left(\frac{4+2Z}{2V}\right)}{\left(\frac{4-Z}{2V}\right)}\Rightarrow Z=1.29$

 mole of B at new Eqm = 4 + 2Z = 6.58