In first order reaction, 20 % concentration remains after 10 min. What would be the rate constant of reaction if lo g 10 ( 5 ) = 0.6989 ?

Question

In first order reaction, 20 % concentration remains after 10 min. What would be the rate constant of reaction if lo g 10 ​ ( 5 ) = 0.6989 ?

TestHub · Accepted Answer

For a first order reaction, the rate constant $k$ is given by the formula:

$k = \frac{2.303}{t} lo g_{10} (\frac{[ A ] _{0}}{[ A ] _{t}})$

Given that $20%$ concentration remains after $10$ min, we have:

$t = 10$ min

$[A]_{t} = \frac{20}{100} [A]_{0} = 0.2 [A]_{0}$

Substituting these values into the formula:

$k = \frac{2.303}{10} lo g_{10} (\frac{[ A ] _{0}}{0.2 [ A ] _{0}})$

$k = \frac{2.303}{10} lo g_{10} (5)$

Given $lo g_{10} (5) = 0.6989$ :

$k = \frac{2.303}{10} \times 0.6989$

$k = 0.2303 \times 0.6989$

$k = 0.1609$ min $^{- 1}$

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