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Question: Trans-1,2 di deutero cyclopropane (A) undergoes a first-order decomposition. The observed rate const...

Trans-1,2 di deutero cyclopropane (A) undergoes a first-order decomposition. The observed rate constant at a certain temperature measured in terms of the disappearance of ‘A’ has 1.52×10−4sec⁡−11.52\times {{10}^{-4}}{{\sec }^{-1}}. Analysis of products showed that the reaction followed two parallel paths, one leading to dideuteropropane (B) and the other to cis-1,2 dideuterocyclopropane (C). (B) was found to constitute 11.2% of the reaction product, independently of the extent of reaction. What is the order of reaction for each path and what is the value of the rate constant for the formation of each of the products?

Explanation

Solution

Calculating the rate constant of B and C will help you to find the order of the reaction.

Complete step by step answer:
-The rate constant of a reaction, denoted by k quantifies the rate of reaction.

-For a hypothetical reaction, in which the chemical species A undergoes one of the two irreversible first-order reactions to form either species B or species C,
A→B+CA\to B+C
A→k1B A→k2C \begin{aligned} & A\xrightarrow{{{k}_{1}}}B \\\ & A\xrightarrow{{{k}_{2}}}C \\\ \end{aligned}

The rate of conversion of A into B is given by
r1(t)=k1[A]=d[A1]dt...(i){{r}_{1}}(t)={{k}_{1}}[A]=\dfrac{d[{{A}_{1}}]}{dt}...(i)
The rate of conversion of A into C is given by
r2(t)=k2[A]=d[A2]dt...(ii){{r}_{2}}(t)={{k}_{2}}[A]=\dfrac{d[{{A}_{2}}]}{dt}...(ii)

The overall rate of reaction for the consumption of A can be written as
r(t)=k[A]=d[A]dt...(iii)r(t)=k[A]=\dfrac{d[A]}{dt}...(iii)

Now, ultimately since each subreaction is assumed to start at the same time, therefore the sum of all rates of appearance of products is equal to the rate of disappearance of A, hence the rates of change in concentration will add up
−d[A]dt=d[A1]dt+d[A2]dt...(iv)-\dfrac{d[A]}{dt}=\dfrac{d[{{A}_{1}}]}{dt}+\dfrac{d[{{A}_{2}}]}{dt}...(iv)
From equation (i), (ii) and (iv) we get,
r(t)=−d[A]dt=r1(t)+r2(t)...(v)r(t)=-\dfrac{d[A]}{dt}={{r}_{1}}(t)+{{r}_{2}}(t)...(v)
Now adding (i) and (ii) and using (v) we get,
r(t)=r1(t)+r2(t) =k1[A]+k2[A] ⇒r(t)=(k1+k2)[A] \begin{aligned} & r(t)={{r}_{1}}(t)+{{r}_{2}}(t) \\\ & ={{k}_{1}}[A]+{{k}_{2}}[A] \\\ & \Rightarrow r(t)=({{k}_{1}}+{{k}_{2}})[A] \\\ \end{aligned}

-According to the question, the compound A is giving rise to B and C products.
B\xleftarrow{11.2%}A\xrightarrow{88.8%}C

-In the case of parallel path reaction,
kB=kA×fractional yield of B =1.52×10−4×0.112 ⇒kB=1.7×10−5sec⁡−1 \begin{aligned} & {{k}_{B}}={{k}_{A}}\times \text{fractional yield of B} \\\ & \text{=1}\text{.52}\times \text{1}{{\text{0}}^{-4}}\times 0.112 \\\ & \Rightarrow {{k}_{B}}=1.7\times {{10}^{-5}}{{\sec }^{-1}} \\\ \end{aligned}
kC=kA×fractional yield of C =1.52×10−4×0.888 ⇒kC=1.35×10−4sec⁡−1 \begin{aligned} & {{k}_{C}}={{k}_{A}}\times \text{fractional yield of C} \\\ & \text{=1}\text{.52}\times \text{1}{{\text{0}}^{-4}}\times 0.888 \\\ & \Rightarrow {{k}_{C}}=1.35\times {{10}^{-4}}{{\sec }^{-1}} \\\ \end{aligned}
Since the rate constant is in sec⁡−1{{\sec }^{-1}}, hence the reaction will be first-order for each path.

Note: When the units of the rate constant are M/s, M/min, M/hr, etc it means it belongs to zero order reaction. When the units of rate constant are s−1,min⁡−1,hr−1,etc.{{s}^{-1}},{{\min }^{-1}},h{{r}^{-1}},etc. , it means it belongs to the first order of reactions. When the units of rate constants are M−1s−1,M−1min⁡−1,M−1hr−1,etc.{{M}^{-1}}{{s}^{-1}},{{M}^{-1}}{{\min }^{-1}},{{M}^{-1}}h{{r}^{-1}},etc. , it means that it belongs to the second-order reactions. When the units of the rate constant are M−2s−1,M−1min⁡−1,M−1hr−1,etc.{{M}^{-2}}{{s}^{-1}},{{M}^{-1}}{{\min }^{-1}},{{M}^{-1}}h{{r}^{-1}},etc. it belongs to third-order reactions.