Integral Method of Analysis (VVBR)

(a) Zero-order Reactions

\[-r_A = k\] \[\begin{align*} \Longrightarrow \quad \frac{C_{A0}}{\varepsilon_A}\frac{d(\ln V)}{dt} &= k \nonumber \\ \frac{C_{A0}}{\varepsilon_A}\int_{\ln V_0}^{\ln V}d(\ln V) &= k\int_0^t dt \nonumber \\ \frac{C_{A0}}{\varepsilon_A}\ln\left(\frac{V}{V_0}\right) &= kt \tag*{(15)}\end{align*}\] Eqn.(15) is plotted in Fig.(13).


(b) First-order Reactions

\[-r_A = kC_A\] \[\begin{align*} \Longrightarrow \quad \frac{C_{A0}}{\varepsilon_A}\frac{d(\ln V)}{dt} &= kC_{A0}\frac{(1-X_A)}{(1+\varepsilon_AX_A)} \nonumber \\ \frac{1}{\varepsilon_A}\frac{1}{V}\frac{dV}{dt} &= \frac{k(1-X_A)}{(1+\varepsilon_AX_A)} \nonumber \\ \text{But, $V=V_0(1+\varepsilon_AX_A) \qquad \Longrightarrow \quad dV=V_0\varepsilon_AdX_A$. So,} \\ \frac{1}{\varepsilon_A}\frac{1}{V_0(1+\varepsilon_AX_A)}(V_0\varepsilon_A)\frac{dX_A}{dt} &= \frac{k(1-X_A)}{(1+\varepsilon_AX_A)} \nonumber \\ \Longrightarrow \quad \int_0^{X_A} \left(\frac{1}{1-X_A}\right)dX_A &= k\int_0^t dt \nonumber \\ \Longrightarrow \quad -\ln(1-X_A) &= kt \nonumber \\ \text{where, for VVBR $X_A = \dfrac{V-V_0}{V_0\varepsilon_A}=\dfrac{\Delta V}{V_0\varepsilon_A}$. Hence,} \\ -\ln\left(1-\frac{\Delta V}{V_0\varepsilon_A} \right) &= kt \tag*{(16)}\end{align*}\]

(c) Second-order Reactions

\[-r_A = kC_A^2\] \[\Longrightarrow \quad \frac{C_{A0}}{\varepsilon_A}\frac{d(\ln V)}{dt} = kC_{A0}^2\frac{(1-X_A)^2}{(1+\varepsilon_AX_A)^2}\] Replacing \(X_A\) with \(V\) and after integration, we get \[\frac{(1+\varepsilon_A)\Delta V}{V_0\varepsilon_A-\Delta V} + \varepsilon_A\ln\left(1-\frac{\Delta V}{V_0\varepsilon_A}\right) = kC_{A0}t \tag*{(17)}\]