Chemistry - Kinetics of a simultaneous parallel radioactive decay

Solution 1:

The question has already been solved by Yashwini and the answer given is correct.$^2$ A more intuitive and specific to question explanation would follow here.

Now, the two reactions given are:

\begin{array}{cc} \require{cancel} \ce{A -> P} &(t_{1/2} = 9\,\mathrm h) \\ \ce{A -> Q} &(t_{1/2} = 4.5\,\mathrm h) \\ \end{array}

Now using the rate law, we get,

\begin{align} -\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm P [A] \tag{1} \\ -\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm Q [A] \tag{2} \\ \end{align}

The rate constant for a first order reaction having a half life of $t_{1/2}$ is defined as:

$$k=\frac{\ln 2}{t_{1/2}} \tag{3}$$

Now, substituting the given values of $t_{1/2}$ into the equations, we get $2k_\mathrm P = k_\mathrm Q$ (since $k\, \alpha \frac{1}{t_{1/2}})$

Now, intuitively since both reactions take place together, it would mean that for every one mole of P formed, two moles of Q forms. Therefore, for every mole of P formed, three moles of A react (since one mole is required for each mole of P and Q).

Now, we add the rate laws ($1$) and $(2)$, since the reactions take place simultaneously, to get:

$$-\frac{\mathrm{d}[A]}{\mathrm{d}t}=(k_\mathrm P +k_\mathrm Q) [A] \tag{4} $$

Now, since using the relation between $k_\mathrm{P}$ and $k_\mathrm{Q}$, we get $k_\mathrm{P} + k_\mathrm{Q} = 3k_\mathrm{P}$

Therefore using the integrated rate law for a first order reaction on equation $(4)$, we get:

$$A=A_0e^{-3k_\mathrm Pt} $$

Now, the amount of $A$ used here would be $A_0 -A$, and we get that value to be:

$$A_\text{used}=A_0\left(1-e^{-3k_\mathrm Pt}\right)$$

Now, as we have previously noted, for every three moles of A used, two moles Q are formed. This means that the amount of Q now in the mixture would be two thirds of $A_\text{used}$. Therefore the amount of Q would be:

$$Q=\frac{2A_0\left(1-e^{-3k_\mathrm Pt}\right)}{3}$$

Now, we are given the condition, $Q = 2A$, substituting values of $Q$ and $A$ into the given relation we get:

$$\begin{align} \frac{\cancel{2A_0}\left(1-e^{-3k_\mathrm Pt}\right)}{3} &= \cancel{2A_0}\left(e^{-3k_\mathrm Pt}\right) \\ \implies 1 -e^{-3k_\mathrm Pt} &= 3e^{-3k_\mathrm Pt} \\ \implies 4e^{-3k_\mathrm Pt} &= 1 \end{align}$$

Solving for $t$, we get:

\begin{align} 3k_\mathrm Pt&=2\ln 2 \\ \\ t&=\frac{2\ln 2}{3k_\mathrm P}\\ \end{align}

Now, using equation $(3)$, we get the rate constant $k_\mathrm P$ to be $\frac{\ln 2}{9}$. Substituting this value into the expression for time, we get:

$$t=\frac{2 \cancel{\ln 2}}{\cancel{3} \frac{\cancel{\ln 2}}{\cancelto{3}{9}}}$$

Therefore, time taken for this condition to happen is:

$$t=2\times 3 = 6\ \mathrm h$$

Solution 2:

Parallel or side reactions of the first order: Concept

$$\require{cancel}\\ \ce{A ->[k_1] B} \ \ t=0\\ \ce{A ->[k_2] C} \ \ t=t$$ $$-\frac{\mathrm d[A]}{\mathrm dt}=k_1[A] + k_2[A] $$ $$-\frac{\mathrm d[A]}{\mathrm dt} = k_\text{eff} [A] \land k_\text{eff}=k_1+k_2$$

Effective order=1

$$\left(t_{1/2}\right)_\text{eff}=\frac {\ln 2}{k_\text{eff}} $$

$$\frac 1 {(t_{1/2})_\text{eff}}=\frac {1}{(t_{1/2})_{1}} + \frac {1} {(t_{1/2})_{2}} $$

$$A_\text{eff}\mathrm e^{-E_\mathrm a/(RT)}=(A_1+A_2)\mathrm e^{-E_\mathrm a/(RT)}$$

Differentiate with regards to $T$,

$${\frac{E_\mathrm a}{RT^2}}\cdot k_\text{eff}=\frac{(E_\mathrm a)_1 k_1}{RT^2}+\frac{(E_\mathrm a)_2 k_2}{RT^2}$$

$$(E_\mathrm a)_\text{eff}=\frac{(E_\mathrm a)_1 k_1 +(E_\mathrm a)_2 k_2}{k_\text{eff}}$$

$$[A]_\mathrm t=[A]_0\mathrm e^{-k_\text{eff}t}$$

$$a_t=a_0\mathrm e^{-(k_1+k_2)t}$$

$$\frac{\mathrm d[B]}{\mathrm dt}=k_1[A]=k_1a_0\mathrm e^{-(k_1+k_2)t}$$

$$\int\limits_{0}^{b_t}\mathrm d[B]=k_1 a_0 \int\limits_0^t\mathrm e^{-(k_1+k_2)t}\,\mathrm dt$$

$$b_t=\frac{k_1 a_0}{-(k_1+k_2)}[\mathrm e^{-(k_1+k_2)t}]_0^t$$

$$b_t=\frac{k_1 a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t}) $$


$$c_t=\frac{k_2 a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t})$$


  • proportion of $B=\frac{[B]}{x}=\frac {k_1}{k_1+k_2}$ [times 100 for percentage]
  • proportion of $C=\frac{[C]}{x}=\frac {k_2}{k_1+k_2}$ [times 100 for percentage]

The actual problem

\begin{align} &\ce{A->[\textit{k}_1]P} &k_1 &= \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{9} \ \text{hr}^{-1} \\ &\ce{A->[\textit{k}_2]Q} &k_2 &= \frac{\ln 2}{t_{1/2}} = \frac{2 \ln2}{9}\ \text{hr}^{-1}\\ \end{align}

$$Q_t=\frac{k_2a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t})=2A_t$$

$$\frac{k_2\cancel{a_0}}{k_1+k_2}\mathrm {(1-e^{-(k_1+k_2)t})}=2\cancel{a_0}\mathrm e^{-(k_1+k_2)t}$$

$$\frac{\cancel 2}{3}(1-\mathrm e^{-k_\text{eff}t})=\cancel 2\mathrm e^{-k_\text{eff}t}$$

$$\mathrm e^{-k_\text{eff}t} = \frac {1} {4}$$

$$\implies k_\text{eff}t = \ln 4 = \frac {3\ln 2}{9} t$$

$$\implies t= 6\mathrm h$$

So that gives the answer as 6 h.