Respuesta :
Option c is the correct answer.
The half-life is defined as the time which is required for an amount to reduce to half of its initial value.
Half life of second order is given by:
[tex] t_\frac{1}{2} [/tex] =[tex] \frac{1}{k[A_o]} [/tex] (1)
where [tex] t_\frac{1}{2} [/tex] = half life
k= rate constant
[tex] A_0 [/tex] = initial concentration
Put the given values of initial concentration and half life in formula (1),we get,
15.4 s = [tex] \frac{1}{k(0.67 m)} [/tex]
k = [tex] \frac{1}{15.4 s\times 0.67 m } [/tex]
k = [tex] \frac{1}{10.318 m s} [/tex]
k = 0.09691 [tex] m^{-1}s^{-1} [/tex]
k = [tex] 9.7 \times 10^{-2} m^{-1} s^{-1} [/tex]
Hence, rate constant of second order reaction is 9.7 \times 10^{-2} m^{-1} s^{-1}.
The rate constant for the second-order decomposition of HIis [tex]\boxed{{\text{c}}{\text{. }}9.7 \times {{10}^{ - 2}}{\text{ }}{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}}}[/tex].
Further Explanation:
Order of reaction:
This determines dependence of rate on power of concentration of reactants involved in any chemical reaction.
A reaction is said to be second-order reaction if its rate varies with square of concentration of one reactant or it can be considered to be dependent on product of concentration of two different reactants.
Decomposition of HI occurs as follows:
[tex]{\text{2HI}} \to {{\text{H}}_2} + {{\text{I}}_2}[/tex]
Half-life is the time period in which half of the radioactive species is consumed. It is written as [tex]{t_{{\text{1/2}}}}[/tex].
Since decomposition of HI is second-order chemical reaction, its half-life can be determined with the help of following formula.
The expression for half-life of a second order reaction is as follows:
[tex]{t_{{\text{1/2}}}} = \dfrac{1}{{k\left[ {{{\text{A}}_{\text{o}}}} \right]}}[/tex] …… (1)
Here,
[tex]{t_{{\text{1/2}}}}[/tex] is half-life period.
k is the rate constant.
[tex]\left[ {{{\text{A}}_{\text{o}}}} \right][/tex] is the initial concentration of reactant.
Rearrange equation (1) for k,
[tex]k = \dfrac{1}{{{t_{{\text{1/2}}}}\left[ {{{\text{A}}_{\text{o}}}} \right]}}[/tex] …… (2)
Substitute 15.4 s for and 0.67 M for [tex]\left[ {{{\text{A}}_{\text{o}}}} \right][/tex] in equation (2).
[tex]\begin{aligned}k &= \frac{1}{{\left( {15.4{\text{ s}}} \right)\left( {0.67{\text{ M}}} \right)}} \\&= 9.7 \times {10^{ - 2}}{\text{ }}{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}} \\\end{aligned}[/tex]
Hence, the correct option is c.
Learn more:
- Rate of chemical reaction: https://brainly.com/question/1569924
- The main purpose of conducting experiments: https://brainly.com/question/5096428
Answer Details:
Grade: Senior School
Subject: Chemistry
Chapter: Chemical Kinetics
Keywords: order, HI, 2HI, H2, I2, second-order reaction, decomposition, half-life, 15.4 s, t1/2, 0.67 M, 9.7*10^-2 M^-1s^-1.
