Question

A first order reaction has a rate constant of 1.15×10^-3s^-1 .how long it will take for 6 gm of reactant to reduce to 2 gm. Also calculate the half life period of this reaction? ​​

Answer 1

Rate. K = 1.15×10^-3s^-1

initial [a] = 6 gm

final [b] = 2 gm

we know that,

$k = \frac{2.303}{t} log \frac{(a)}{(b)} \\ \\ t = \frac{2.303}{k} log \frac{(a)}{(b)} \\ \\ t = \frac{2.303}{1.15 \times {10}^{ - 3} } log \frac{(6)}{(2)} \\ \\ t = \: 2.00 \times \: log(3) \\ \\ t = 2 \times 0.477121255 \\ \\ t = 0.95424251 \: seconds$

t = 0.95424251 seconds

now for,

Half Life period

t1/2 = 0.693/k

t1/2 = 0.693/1.15×10^-3

t1/2 = 0.6026 × 10^3

I hope the answer is true...

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Hamesha muskurate rahiye ☺️