Question

For a first order reaction if time taken for 50% completion is 10
sec then time required for 90% completion of the reaction is

21.25 sec
46.25 sec
33.23 sec
51.51 sec​

Answer 1

Answer:

Time = 33.23 s

Explanation:

Given:

• The reaction is a first order reaction

• Time taken for 50% completion = 10 s

To Find:

• Time required for 90% completion

Solution:

Let initial concentration be R₀ and final concentration be R

By given,

R = 50/100 × R₀ = 0.5 R₀

First finding the rate constant of the reaction,

For a first order reaction rate constant is given as,

$\sf k=\dfrac{2.303}{t} \: log\: \dfrac{[R_0]}{R}$

Substitute the data,

$\sf k=\dfrac{2.303}{10} \: log\: \dfrac{R_0}{0.5\times R_0}$

$\sf k=\dfrac{2.303}{10} \: log\: \dfrac{1}{0.5}$

$\sf k=0.2303\times (1-0.69897)$

$\sf k=0.2303\times 0.30103$

$\sf k =0.0693272\: s^{-1}$

Now finding the time required to complete 90% of the reaction,

Final concentration R =  R₀ - 0.9 R₀ = 0.1 R₀

Substitute in the above formula,

$\sf 0.0693272=\dfrac{2.303}{t} \: log\dfrac{R_0}{0.1\:R_0}$

$\sf 0.0693272=\dfrac{2.303}{t} \: log\dfrac{1}{0.1}$

$\sf 0.0693272=\dfrac{2.303}{t} \times 1$

$\sf t=\dfrac{2.303}{0.0693272}$

$\sf t=33.23\: s$

Hence to complete 90% of the reaction the time required is 33.23 s

Answer 2

${\bf{Given\::}}$

• A first order reaction takes 10 second for 50% completion.

$\\ {\bf{To\: Find\::}}$

• Time required for 90% completion of the reaction.

$\\ {\bf{Solution\::}}$

As we know that,

➣ Rate constant (K) for first order reaction is given as,

$\orange\bigstar\:{\Large\mid}\:\bf\purple{K\:=\:\dfrac{2.303}{t}\:\log\dfrac{[R_o]}{[R]}\:}\:{\Large\mid}\:\green\bigstar$

Where,

• t = time taken for 50% completion, i.e. 10 second

• [R]₀ = Initial concentration of reactant.

• [R] = Final concentration of reactant.

Let,

• [R]₀ = 100

• [R] = 100 - 50 = 50 [due to 50% completion]

➠ $\tt{K\:=\:\dfrac{2.303}{10}\:\log\dfrac{100}{50}\:}$

➠ $\tt{K\:=\:0.2303\times{\log{2}}\:}$

➠ $\tt{K\:=\:0.2303\times{0.301}\:}$

➠ $\bf{K\:=\:0.0693\:sec^{-1}\:}$

Now,

➣ For same reaction the time required for 90% completion of the reaction is,

Here,

• [R] = 100 - 90 = 10 [due to 90% completion]

$\dashrightarrow\:\tt{0.0693\:=\:\dfrac{2.303}{t'}\:\log\dfrac{100}{10}\:}$

$\dashrightarrow\:\tt{0.0693\:=\:\dfrac{2.303}{t'}\times{\log{10}}\:}$

$\dashrightarrow\:\tt{0.0693\:=\:\dfrac{2.303}{t'}\times{1}\:}$

$\dashrightarrow\:\tt{0.0693\:=\:\dfrac{2.303}{t'}\:}$

$\dashrightarrow\:\tt{t'\:=\:\dfrac{2.303}{0.0693}\:}$

$\dashrightarrow\:\bf\pink{t'\:=\:33.23\: second}$

∴ ⑶ Time required for 90% completion of the reaction is 33.23 second.