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  •  In a first order reaction, 75% of the reactants disappeared in 1.386 hr. The value of the rate constant is:  Option: 1</

 In a first order reaction, 75% of the reactants disappeared in 1.386 hr. The value of the rate constant is:

 

Option: 1

0.693 \mathrm{hr}^{-1} \\


Option: 2

1 \mathrm{hr}^{-1} \\


Option: 3

1.386 \mathrm{hr}^{-1} \\


Option: 4

2.772 \mathrm{hr}^{-1}


Answers (1)

best_answer

Time taken for completion of 75 % of the reaction is double than that of the time taken for half of the reaction to complete.
So,  \mathrm{t_{75 \%}-2 \times t_{\frac{1}{2}}}
But.

\mathrm{t_{75 %} =1.386 h r }
\mathrm{\therefore t_{\frac{1}{2}} =\frac{1.386}{2}=0.693 h r }

Now,
\mathrm{ k=\frac{\ln 2}{t_{\frac{1}{2}}}=\frac{\ln 2}{0.693} \approx 1 h r^{-1} }.

Posted by

Sanket Gandhi

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