Chemistry - Chemical Kinetics Question with Solution | TestHub

At time $t=0$,

equivalent of X reacting with A = equivalent of A present.

$(M_a)V_0=n_0\alpha \dots \dots (i)$

At time $t=t$,

equivalent of X reacting with B = equivalent of B present.

$(M_b)V_B=x\beta .\dots (ii)$

$V_B=\frac{x\beta }{(M_b)}.\dots \left(\mathrm{i}i\right)$ equivalent of X reacting with A = equivalent of A present at $t=t$.

$(V_A)(M_a)=(n_0-x)\alpha$ Let $V_A=V_0-y$.

$M_a(V_0-y)=n_0\alpha -x\alpha ..\dots (iii)$

From (i) and (iii):

$M_{a}y=x\alpha$

$x=\frac{M_{a}y}{\alpha }.\dots (iv)$

From (iv) and (ii):

$V_B=\frac{M_{a}y}{\alpha }\frac{\beta }{M_b}=\frac{a\beta }{b\alpha }y$

At $t=\infty$, $y=V_0$. So,

$V_B=\frac{a\beta }{b\alpha }{V}_0$