Answer
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Hint: Here decomposition means breaking of\[S{{O}_{2}}C{{l}_{2}}\] is a first order means its rate of decomposition depends on single power of concentration.
Step by step solution:
Decomposition reaction of \[S{{O}_{2}}C{{l}_{2}}\]
\[S{{O}_{2}}C{{l}_{2}}(g)\to S{{O}_{2}}\left( g \right)\text{ }+\text{ }C{{l}_{2}}\left( g \right)\]
We know that for a 1st order reaction,
\[r=k[C]\]
Where r is rate of reaction,
‘k’ is the rate constant
‘C’ is the concentration of \[S{{O}_{2}}C{{l}_{2}}\]
And we know that for 1st order reaction:
\[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }=\text{ }\dfrac{0.693}{k}~\]
Where \[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }\]is half life of reaction.
It is given that \[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }\]= 60 min
Now we put this given value of half life in above formula and rearranging:
\[\text{k }=\text{ }\dfrac{0.693}{60}~\]
= 0.693 / 6
k= 0.01155 \[{{\min }^{-1}}~\]
k= 1.1925 \[\times {{10}^{-4}}{{\sec }^{-1}}\]
Alternative method
We can also solve this problem by using the first order reaction equation, which is
\[k=\dfrac{2.303}{t}\log \dfrac{[{{C}_{0}}]}{[C]}\]
And we know that at \[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }\]value of [C] = \[[{{C}_{0}}]\]/2
Now we will put value of \[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }\]value of [C] in above equation:
\[k=\dfrac{2.303}{60}\log \dfrac{[{{C}_{0}}]}{(\dfrac{[{{C}_{0}}]}{2})}\]
k= 0.01155 \[{{\min }^{-1}}~\]
k= 1.1925 \[\times {{10}^{-4}}{{\sec }^{-1}}\]
So, from the above calculation we can see that the correct answer is option “D”.
Note: Options can be both in units of minutes or seconds. But in this question options are given in a minute unit so no need to convert the value of ‘k’ in seconds. From units of ‘k’ we can also check the order of reaction.
Step by step solution:
Decomposition reaction of \[S{{O}_{2}}C{{l}_{2}}\]
\[S{{O}_{2}}C{{l}_{2}}(g)\to S{{O}_{2}}\left( g \right)\text{ }+\text{ }C{{l}_{2}}\left( g \right)\]
We know that for a 1st order reaction,
\[r=k[C]\]
Where r is rate of reaction,
‘k’ is the rate constant
‘C’ is the concentration of \[S{{O}_{2}}C{{l}_{2}}\]
And we know that for 1st order reaction:
\[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }=\text{ }\dfrac{0.693}{k}~\]
Where \[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }\]is half life of reaction.
It is given that \[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }\]= 60 min
Now we put this given value of half life in above formula and rearranging:
\[\text{k }=\text{ }\dfrac{0.693}{60}~\]
= 0.693 / 6
k= 0.01155 \[{{\min }^{-1}}~\]
k= 1.1925 \[\times {{10}^{-4}}{{\sec }^{-1}}\]
Alternative method
We can also solve this problem by using the first order reaction equation, which is
\[k=\dfrac{2.303}{t}\log \dfrac{[{{C}_{0}}]}{[C]}\]
And we know that at \[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }\]value of [C] = \[[{{C}_{0}}]\]/2
Now we will put value of \[{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}~\text{ }\]value of [C] in above equation:
\[k=\dfrac{2.303}{60}\log \dfrac{[{{C}_{0}}]}{(\dfrac{[{{C}_{0}}]}{2})}\]
k= 0.01155 \[{{\min }^{-1}}~\]
k= 1.1925 \[\times {{10}^{-4}}{{\sec }^{-1}}\]
So, from the above calculation we can see that the correct answer is option “D”.
Note: Options can be both in units of minutes or seconds. But in this question options are given in a minute unit so no need to convert the value of ‘k’ in seconds. From units of ‘k’ we can also check the order of reaction.
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