Question

In the first order reaction, 75% of the reactant gets disappeared in 1.386 h. The rate constant of the reaction is

Answer 1

The rate constant of the given reaction is $\sf2.8\times10^{-4}s^{-1}$

Step by step Explanation :

Given:

In the first order reaction 75% of the reactant disappear in 1.386hr

To Find :

The rate constant of the reaction

Theory :

• First order reaction :

The rate of reaction depends only on one concentration term of reactant.

• Calucation of Rate Constant

$\bf\blue{\log_e(a)-\log_e(a-x)=kt}$

where ,

• ' a ' is the concentration of reactant at t = 0
• (a-x) is the concentration of reactant after time t .
• k = Rate Constant

Solution :

Let the initial concentration of the reaction be a .

75% of the reaction gets disappeared in t = 1.386hr

Then , Concentration of the reactant after t = 1.386 hr is :

$\sf=a-\dfrac{75}{100}\times\:a$

$\sf=a-\dfrac{3}{4}$

$\sf=\dfrac{4a-3a}{4}$

$\sf=\dfrac{a}{4}$

We have to find the Rate Constant of the given reaction.

For First order Reaction , Rate Constant:

$\bf\:k=\log_e(\dfrac{a}{a-x})\times\dfrac{1}{t}$

$\sf\implies\:k=\dfrac{2.303}{t}\times\log_{10}(\dfrac{a}{a/4})$

$\sf\implies\:k=\dfrac{2.303}{t}\times\log_{10}(4)$

Put the given values

$\sf\implies\:k=\dfrac{2.303}{1.386\times60\times60}\times\log_{10}(4)$

$\sf\implies\:k=\dfrac{2.303}{1.386\times60\times60}\times0.602$

$\sf\implies\:k=\dfrac{2.303}{49.89}\times0.602\times10^{-2}$

$\sf\implies\:k=0.0461\times0.602\times10^{-2}$

$\sf\implies\:k=0.0461\times0.602\times10^{-2}$

$\sf\implies\:k=0.0277\times10^{-2}$

$\sf\implies\:k=0.028\times10^{-2}$

$\sf\implies\:k\approx2.8\times10^{-4}s^{-1}$

Therefore , the rate constant of the given reaction is $\sf2.8\times10^{-4}s^{-1}$

Answer 2

The specific rate constant (k) is the proportionality constant relating the rate of the reaction to the concentrations of reactants. The rate law and the specific rate constant for any chemical reaction must be determined experimentally. The value of the rate constant is temperature dependent.