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  • Consider the following parallel reactions being given by A (t1/2 = 1..386  102 

Consider the following parallel reactions being given by A (t1/2 = 1..386 \times 102 hours), each path being first order.


If the distribution of B in the product mixture is 50%, the partial half life (in hours) of A for conversion into B is:

Option: 1

231


Option: 2

131


Option: 3

115.5


Option: 4

31


Answers (1)

best_answer

As we have learned,

 

 

Parallel First Order Kinetics -

In this situation, B and C both are forming. These types of reactions are known as parallel reactions. Both these reactions are first order reactions with rate constants K1 and K2 respectively and half-lives as t(1/2)1 and t(1/2)2.

For these parallel reactions, we need to find:

  • Effective order
  • Effective rate constant
  • Effective t1/2
  • Effective Activation energy
  • [A], [B], [C] with time (t) variation
  • % of [B] and % of [C]

We know that the rate equations are given as follows:

\\\mathrm{r_{1}\:=\: \frac{-dA}{dt}\:=\: K_{1}[A] }\\\\\mathrm{r_{2}\: =\: \frac{-dA}{dt}\: =\: K_{2}[A]}

\\\mathrm{Thus,\: overall\: rate\: of\: reaction\: is:}\\\\\mathrm{\frac{-dA}{dt}\: =\: K_{1}[A]\: +\: K_{2}[A]\: =\: (K_{1}\: +\: K_{2})[A]}

\mathrm{Thus,\: rate\: =\: (K_{1}\: +\: K_{2})[A]^{1}}

\mathrm{\mathbf{Effective\: Rate\: Constant(K_{eff})}\: =\: (K_{1}\: +\: K_{2})}

\mathrm{\mathbf{Effective\: order\: of\: reaction}\: =\: 1}

\mathrm{Now, effective\: half-life(t_{1/2})\:=\: \frac{0.693}{K_{eff}}\: =\: \frac{0.693}{K_{1}\: +\: K_{2}}}

                                                           \mathrm{\Rightarrow\: \frac{0.693}{\frac{0.693}{(t_{1/2)_{1}}}+\frac{0.693}{(t_{1/2})_{2}}}}

\\\mathrm{Thus,\: effective\: half\: life\: is\: given\: as:}\\\\\mathrm{\frac{1}{(t_{1/2})_{eff}}\: =\: \frac{1}{(t_{1/2})_{1}}\: +\:\frac{1}{(t_{1/2})_{2}} }

NOTE: Effective activation energy, [A], [B], [C] with time (t) variation and % of [B] and % of [C] will be discussed in later concepts.

-

 

 

\begin{array}{l}{\frac{2 k_{1}}{2 k_{1}+3 k_{2}}=0.5 \text { and }} {k_{1}+k_{2}=\frac{0.693}{1.386 \times 10^{2}}=5 \times 10^{-3} \mathrm{\: h}^{-1}}\end{array}
\begin{array}{l}{\text { Solving } \frac{k_{1}}{k_{2}}=\frac{2}{3} \text { and } k_{1}=2 \times 10^{-3} \mathrm{\: h}^{-1} \text { and }} \\\\ {\: k_{2}=3 \times 10^{-3} \mathrm{\: h}^{-1}} \\\\ {t_{1 / 2(\mathrm{A} \rightarrow \mathrm{B})}=\frac{0.693}{k_{2}}=\frac{0.693}{3 \times 10^{-3}}=231 \mathrm{\: h}}\end{array}
 

Posted by

Pankaj Sanodiya

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