Chemistry, 12.07.2019 17:30 bxbykyah

The half-life for the second-order decomposition of hi is 15.4 s when the initial concentration of hi is 0.67 m. what is the rate constant for this reaction? a. 1.0 x 10-2 m-1s-1 b. 4.5 x 10-2 m-1s-1 c. 9.7 x 10-2 m-1s-1 d. 2.2 x 10-2 m-1s-1 e. 3.8 x 10-2 m-1s-1

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Answer from: ladnerhailey16

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:

$t_\frac{1}{2}$ = $\frac{1}{k[A_o]}$ (1)

where $t_\frac{1}{2}$ = half life

k= rate constant

A_0 = initial concentration

Put the given values of initial concentration and half life in formula (1),we get,

15.4 s = $\frac{1}{k(0.67 m)}$

k = $\frac{1}{15.4 s\times 0.67 m }$

k = $\frac{1}{10.318 m s}$

k = 0.09691 $m^{-1}s^{-1}$

k = $9.7 \times 10^{-2} m^{-1} s^{-1}$

Hence, rate constant of second order reaction is 9.7 \times 10^{-2} m^{-1} s^{-1}.

Answer from: blakeley7785

The rate constant for the second-order decomposition of HIis $\boxed{{\text{c}}{\text{. }}9.7 \times {{10}^{ - 2}}{\text{ }}{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}}}$ .

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:

${\text{2HI}} \to {{\text{H}}_2} + {{\text{I}}_2}$

Half-life is the time period in which half of the radioactive species is consumed. It is written as ${t_{{\text{1/2}}}}$ .

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:

${t_{{\text{1/2}}}} = \dfrac{1}{{k\left[ {{{\text{A}}_{\text{o}}}} \right]}}$ …… (1)

Here,

${t_{{\text{1/2}}}}$ is half-life period.

k is the rate constant.

$\left[ {{{\text{A}}_{\text{o}}}} \right]$ is the initial concentration of reactant.

Rearrange equation (1) for k,

$k = \dfrac{1}{{{t_{{\text{1/2}}}}\left[ {{{\text{A}}_{\text{o}}}} \right]}}$ …… (2)

Substitute 15.4 s for and 0.67 M for $\left[ {{{\text{A}}_{\text{o}}}} \right]$ in equation (2).

$\begin{aligned}k &= \frac{1}{{\left( {15.4{\text{ s}}} \right)\left( {0.67{\text{ M}}} \right)}} \\&= 9.7 \times {10^{ - 2}}{\text{ }}{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}} \\\end{aligned}$

Hence, the correct option is c.

Learn more:

Rate of chemical reaction: The main purpose of conducting experiments:

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.

Answer from: Quest

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Answer from: Quest

am i too late?

explanation:

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