Chapter 13, Problem 13.65QP

In the presence of excess thiocyanate ion, SCN⁻, the following reaction is first order in chromium(III) ion, Cr³⁻; the rate constant is 2.0 × 10⁻⁶/s.

${\text{Cr}}^{3+}(aq)+{\text{SCN}}^{-}(aq)\stackrel{}{\to }\text{Cr}{(\text{SCN})}^{2+}(aq)$

What is the half-life in hours? How many hours would be required for the initial concentration of Cr³⁻ to decrease to each of the following values: 25.0% left, 12.5% left, 6.25% left, 3.125% left?

Interpretation Introduction

Interpretation:

The half-life of ${\text{SCN}}^{\text{-}}$ and the time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{25\%}$ , $\text{12}\text{.5\%}$, $\text{3}\text{.125\%}$  and $\text{6}\text{.5\%}$ of its initial value has to be calculated.

Concept Introduction:

Half life period:

The time taken by the concentration of reaction to get reduced of its original concentration is called as half-life reaction.

Half life period can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0693}}{\text{k}}$

The half-life period of substance is related to rate constant but it is independent of concentration of reactants.

Answer to Problem 13.65QP

The half life of ${\text{SCN}}^{\text{-}}$ is $\text{3}{\text{.465×10}}^{\text{5}}\text{sec(96}\text{.25 or 96}\text{hr)}$.

The time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{25\%}$ of its initial value is $\text{1}{\text{.9×10}}^{\text{2}}\text{hr}$.

The time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{12}\text{.5\%}$ of its initial value is $\text{2}{\text{.9×10}}^{\text{2}}\text{hr}$.

The time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{6}\text{.5\%}$ of its initial value is $\text{3}{\text{.9×10}}^{\text{2}}\text{hr}$.

The time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{3}\text{.125\%}$ of its initial value is $\text{4}{\text{.8×10}}^{\text{2}}\text{hr}$.

Explanation of Solution

To calculate the half-life of ${\text{SCN}}^{\text{-}}$

Substitute $\text{k=9}\text{.2/s}$

Half life period can be calculated using the equation,

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0693}}{\text{k}}$

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{2}{\text{.0×10}}^{\text{-6}}\text{/s}}$

${\text{t}}_{\text{1/2}}\text{=3}{\text{.465×10}}^{\text{5}}\text{sec(96}\text{hr)}$

The half life of ${\text{SCN}}^{\text{-}}$ = $\text{3}{\text{.465×10}}^{\text{5}}\text{sec(96}\text{.25 or 96}\text{hr)}$.

To calculate time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{25\%}$ of its initial value

${\text{t}}_{\text{25\%}\text{left}}{\text{=t}}_{\text{1/4}}\text{=2×(96}\text{.25}\text{hr)=192}\text{.5=1}{\text{.9×10}}^{\text{2}}\text{hr}$

For the concentration to decrease by $\text{25\%}$ , it takes $\text{1}{\text{.9×10}}^{\text{2}}\text{hr}$.

To calculate time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{12}\text{.5\%}$ of its initial value

${\text{t}}_{\text{12}\text{.5\%}\text{left}}{\text{=t}}_{\text{1/8}}\text{=3×(96}\text{.25}\text{hr)=288}\text{.75=2}{\text{.9×10}}^{\text{2}}\text{hr}$

For the concentration to decrease by $\text{12}\text{.5\%}$ , it takes $\text{2}{\text{.9×10}}^{\text{2}}\text{hr}$.

To calculate time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{6}\text{.5\%}$ of its initial value

${\text{t}}_{\text{6}\text{.25\%}\text{left}}{\text{=t}}_{\text{1/16}}{\text{=4×t}}_{\text{1/2}}\text{=4(96}\text{.25}\text{hr)=385}\text{.0=3}{\text{.9×10}}^{\text{2}}\text{hr}$

For the concentration to decrease by $\text{6}\text{.5\%}$ , it takes $\text{3}{\text{.9×10}}^{\text{2}}\text{hr}$.

To calculate time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease $\text{3}\text{.125\%}$ of its initial value

${\text{t}}_{\text{3}\text{.125\%}\text{left}}{\text{=t}}_{\text{1/32}}{\text{=5×t}}_{\text{1/2}}\text{=4(96}\text{.25}\text{hr)=481}\text{.25=4}{\text{.8×10}}^{\text{2}}\text{hr}$

For the concentration to decrease by $\text{3}\text{.125\%}$ , it takes $\text{4}{\text{.8×10}}^{\text{2}}\text{hr}$.

Conclusion

The half-life of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ and the time duration for decrease in concentration of ${\text{SCN}}^{\text{-}}$ to decrease to $\text{25\%}$ , $\text{12}\text{.5\%}$, $\text{6}\text{.5\%}$ and $\text{3}\text{.125\%}$ to its initial value was calculated.