Question

30% of a 1st order reaction completes in 200 days. how much time will it require for 80% completion

Answer 1

Answer:

The reaction will require 900 days for 80% completion.

Explanation:

For first-order reaction we have,

$k=\frac{2.303}{t}log\frac{A_0}{A}$      (1)

Where,

k=rate constant

t=time during which the reaction occurs

A₀=initial concentration of the reactant

A=final concentration of the reactant

For 30%completion

Let [A₀]=100

If 30% of the reaction is completed then [A]=70

Time required for the reaction to proceed(t)=200 days

Then equation (1) can be written as,

$k=\frac{2.303}{200}log\frac{100}{70}$      (2)

For 80% completion

If 80% of the reaction is completed then [A]=20

So equation (1) can be written as,

$k=\frac{2.303}{t}log\frac{100}{20}$     (3)

Using equation (2) in equation (3) we get;

$\frac{2.303}{t}log\frac{100}{20}=\frac{2.303}{200}log\frac{100}{70}$

$\frac{1}{t}log 5=\frac{1}{200}log\frac{100}{70}$

$\frac{1}{t}\times 0.698=\frac{1}{200}(log100-log70)$

$\frac{1}{t}\times 0.698=\frac{1}{200}(2-1.845)$

$\frac{1}{t}\times 0.698=\frac{1}{200}\times 0.155$

$t=200\times \frac{0.698}{0.155}$

$t\approx 900\ days$

Hence, the reaction will require 900 days for 80% completion.

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