Question

The relationship between half life of a reaction and the order of reaction is:
a.) \[{{t}_{\dfrac{1}{2}}}\propto \dfrac{1}{{{a}^{(n+1)}}}\]
b.) \[{{t}_{\dfrac{1}{2}}}\propto \dfrac{1}{{{a}^{(n+2)}}}\]
c.) \[{{t}_{\dfrac{1}{2}}}\propto \dfrac{1}{{{a}^{(n)}}}\]
d.) \[{{t}_{\dfrac{1}{2}}}\propto \dfrac{1}{{{a}^{(n-1)}}}\]

Answer 1

Hint: Here the question is asking for the relationship between half life and order of reaction for this we have to write the relationship between rate of reaction and concentration and then we will put conditions of half life.


Complete step by step solution:
We know that the half-life of the reaction is the time interval in which the initial concentration of the reactants is reduced to half the value. The half-life reaction depends on the order of the reaction and takes different forms for different reaction orders.
 Let the order of reaction is ‘n’, concentration is ‘C’ and ‘k’ is the rate constant.
So, the rate of reaction \[r=k{{[C]}^{n}}\]
And we can write \[r=-\dfrac{dC}{dt}\]
Now putting this value of ‘r’ in above reaction:
\[-\dfrac{dC}{dt}=k{{[C]}^{n}}\]
By rearranging:
\[-\dfrac{dC}{{{[C]}^{n}}}=kdt\]
Now, we will integrate both side and take limits C = \[{{C}_{\circ }}\](initial concentration) at t = 0 and C = C at t = t
\[-\int_{{{C}_{\circ }}}^{C}{\dfrac{dC}{{{[C]}^{n}}}}=k\int_{0}^{t}{dt}\]
 So, after integration:
\[-\left[ \dfrac{{{C}^{-n+1}}}{-n+1}-\dfrac{{{C}_{\circ }}^{-n+1}}{-n+1} \right]=k[t-0]\]
Now we will put conditions of half life:
\[t={{t}_{\dfrac{1}{2}}},C=\dfrac{{{C}_{\circ }}}{2}\], where \[{{t}_{\dfrac{1}{2}}}\]is half life
\[{{t}_{\dfrac{1}{2}}}=\dfrac{{{C}_{\circ }}^{-n+1}{{2}^{n-1}}}{(-n+1)k}\]
So,
\[{{t}_{\dfrac{1}{2}}}\propto \dfrac{1}{{{C}_{\circ }}^{n-1}}\]
Here, \[{{C}_{\circ }}\]= a
Then, \[{{t}_{\dfrac{1}{2}}}\propto \dfrac{1}{{{a}^{n-1}}}\]
So, the correct answer is “Option D”.

Note: Here rate of reaction we will write in terms of reactant. And if you see the dependency of half life in order of reaction then there are some other terms also there.