A first order reaction was started with a decimolar solution of t… - Vidyadip

MCQ

A first order reaction was started with a decimolar solution of the reactant, 8 minutes and 20 seconds later its concentration was found to be $M/100$. So the rate of the reaction is

A
$2.303 \times {10^{ - 5}}\;{\sec ^{ - 1}}$
B
$2.303 \times {10^{ - 4}}\;{\sec ^{ - 1}}$
✓
$4.606 \times {10^{ - 3}}\;{\sec ^{ - 1}}$
D
$2.606 \times {10^{ - 5}}\;{\sec ^{ - 1}}$

✓

Answer

Correct option: C.

$4.606 \times {10^{ - 3}}\;{\sec ^{ - 1}}$

c
For first order reaction

$K = \frac{{2.303}}{t}\log \frac{a}{{a - x}}$

Given:

$a = \frac{1}{{10}} = .1\,m$;

$a - x = \frac{1}{{100}} = .01\,m$;

$ t = 500 $ $\sec$

$\therefore \;\;K = \frac{{2.303}}{{500}}\log \frac{{.10}}{{.01}} = \frac{{2.303}}{{500}}\log \,10$

$ = \frac{{2.303}}{{500}} = 0.004606 = 4.6 \times {10^{ - 3}}\,{\sec ^{ - 1}}$.

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