Question

the rate constant for a second order reaction is 3.33*10^-2 L mol^-1s^-1.the initial concentration of the reactant is 0.5 mol /L. calculate the half life of the reaction
A-0.95 min, B- 1.95min, C- 10min,D- 3.45 min​

Answer 1

Answer:

0.95 min

Explanation:

t half for second order+ 1/(k*[INITIAL CONCENTRATION]=1/(0.033*0.5)=60.60SECOND   IN SECONDS

=1/(0.033*0.5)*60

=1.01 min

= closest  equal 0.95 min

Answer 2

Answer:

For a second order reaction , the half life period will be equal to 1 min.

Therefore, the option (A) has closest value to it.

Explanation:

For second order reaction,

Given, the rate constant of the reaction, $K=3.33*10^{-2}Lmol^{-1}s^{-1}$

The initial concentration of reaction [A₀] = 0.5molL⁻¹

Half life period is the time required for the concentration of the reactant to reduce to half of its original value.

Half life period of second order reaction is inversely proportional to the initial concentration of reactants.

Half life period of 2nd order reaction, $t_{\frac{1}{2} }=\frac{1}{k[A_{o}]}$

$t_{\frac{1}{2} }=\frac{1}{( 3.33*10^{-2} L mol^{-1}s^{-1})(0.5molL^{-1})}$

$t_{\frac{1}{2} }=60sec$

$t_{\frac{1}{2} }=1min$

Therefore, the half life of the given 2nd order reaction is equal to 1min.