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A reaction follows first order kinetics having rate constant \left ( K \right ) is 2 \times 10^{-2} s^{-1} at 50^{\circ} \mathrm{C}, what will be the half-life of the reaction at 60^{\circ} \mathrm{C} ? (Assume E_{a}=60 \mathrm{~kJ} / \mathrm{mol} and the pre-exponential factor \mathrm{A} is (instant).

Option: 1

15 seconds


Option: 2

18 seconds

 


Option: 3

21 seconds


Option: 4

24 seconds


Answers (1)

best_answer

Given that, \mathrm{K} at 50^{\circ} \mathrm{C} is 2 \times 10^{-2} \mathrm{~s}^{-1},

To find t_{1/2} at 60^{\circ} \mathrm{C}, so first we have to Calculate rate constant at 60^{\circ} \mathrm{C}.

So, A/c to Arrhenius Equation

\mathrm{\log \left(\frac{k_{2}}{k_{1}}\right) =\frac{\mathrm{Fa}_{a}}{2.303 \mathrm{R}}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] }
\mathrm{\log \left(\frac{k_{2}}{2 \times 10^{-2}}\right) =\frac{60 \times 10^{3}}{2.303 \times 8.314} \times\left[\frac{1}{323}-\frac{1}{333}\right] }
\mathrm{\log \left(\frac{k_{2}}{2 \times 10^{-2}}\right) =0.291 }

\mathrm{k_{2} =2 \times 10^{-2} \text { Antilog }(0.291)}
\mathrm{k_{2} =2 \times 10^{-2} \times 1.954=0.039 \mathrm{~s}^{-1}}

Now, we con calculate the half-like \mathrm{\left(t_{1 / 2}\right) \, at \, 60^{\circ} \mathrm{c}} using the rate constant \mathrm{k_{2}}.

\mathrm{t_{1 / 2}=\frac{0.693}{k_{2}}=\frac{0.693}{0.039} =17.76 \text { seconds }}
                                              \mathrm{=18 \text { seconds }}.

Posted by

Devendra Khairwa

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