A first order reaction is 50 \% completed in 1.26 × 10^ 14 ~s . How much time would it takes for 100\% completion?

Correct answer: D. Infinite k -for first order. 0.693 t_ 1/2 = 0.693 1.26 × 10^ 14 for 100 \% completion. t=(2.303)/(k) (a)/(a-x) t= 2.303 × 1.26 × 10^ 14 0.693 (100)/(0) t=∞

JIPMER 2017

Exam: JIPMER
Year: 2017
Subject: Chemistry
Chapter: Chemical Kinetics

JIPMER 2017 · Chemistry · Chemical Kinetics

A first order reaction is 50 \% completed in 1.26 × 10^ 14 ~s . How...

A first order reaction is

50 \%

completed in

1.26 \times 10^{14} \mathrm{~s}

. How much time would it takes for 100\% completion?

Options

A $1.26 \times 10^{15} \mathrm{~s}$
B $2.52 \times 10^{14} \mathrm{~s}$
C $2.52 \times 10^{28} \mathrm{~s}$
DInfinite

Correct answer

D. Infinite

Step-by-step solution

k

-for first order.

\frac{0.693}{t_{1/2}}=\frac{0.693}{1.26 \times 10^{14}}

for

100 \%

completion.

\because t=\frac{2.303}{k} \log \frac{a}{a-x}

t=\frac{2.303 \times 1.26 \times 10^{14}}{0.693} \log \frac{100}{0}

\therefore t=\infty

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