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  • Under the same reaction conditions, initial concentration of <img alt="2.345 \; \mathrm{moldm}^{-3}" src="https://entrancecorner.oncodecogs.com/gif.latex?2.345%20%5C%3B%20%5Cmathrm%7Bmoldm%7D%5E%7B

Under the same reaction conditions, initial concentration of 2.345 \; \mathrm{moldm}^{-3} of a substance becomes half in 1 / 60 \mathrm{hr}$ and $1 / 45 \mathrm{hr} through first order and zero order kinetics respectively. Ratio \left(k_1 / k_0\right) of the rate constants for first order \left(k_1\right) and zero order \left(k_0\right) of the reaction is

 

Option: 1

0.788


Option: 2

1.2389


Option: 3

8.906


Option: 4

5.981


Answers (1)

best_answer

Given information:

Initial concentration [A]_0=2.345 \mathrm{~mol} / \mathrm{dm}^3

Half-life for first order kinetics t_1=60 seconds

Half-life for zero order kinetics t_0=80 seconds

For first order kinetics, we have the equation:

\ln \left(\frac{[A]_t}{[A]_0}\right)=-k_1 \cdot t_1
Substituting the values, we get:
\ln \left(\frac{0.5}{[A]_0}\right)=-k_1 \cdot 60
Simplifying, we have:

\ln (0.5)=-60 k_1

For zero order kinetics, we have the equation:

[A]_t=[A]_0-k_0 \cdot t_0

Substituting the values, we get:

0.5[A]_0=[A]_0-k_0 \cdot 80
Simplifying, we have:

0.5[A]_0=[A]_0-80 k_0
Rearranging the equation, we get:

\begin{gathered} 80 k_0=[A]_0-0.5[A]_0 \\ 80 k_0=0.5[A]_0 \\ k_0=\frac{[A]_0}{160} \end{gathered}

Now let's substitute the given initial concentration [A]_0=2.345 \mathrm{~mol} / \mathrm{dm}^3:

\begin{aligned} & k_0=\frac{2.345}{160} \\ & k_0=0.01465625 \mathrm{~mol} /\left(\mathrm{dm}^3 \cdot \mathrm{s}\right) \end{aligned}

To find k_1, we rearrange the equation from first order kinetics:

\ln (0.5)=-60 k_1
Solving for k_1 :
\begin{aligned} & k_1=-\frac{\ln (0.5)}{60} \\ & k_1=0.0115518 \mathrm{~s}^{-1} \end{aligned}
Now we can find the ratio
\left(\frac{k_1}{k_0}\right)=\frac{0.0115518 \mathrm{~s}^{-1}}{0.01465625 \mathrm{~mol} /\left(\mathrm{dm}^3 \cdot \mathrm{s}\right)}\left(\frac{k_1}{k_0}\right) \approx 0.788

Therefore, the corrected ratio of the rate constants \left(\frac{k_1}{k_0}\right) for first order and zero order kinetics is approximately 0.788.

Posted by

Ritika Jonwal

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