Chapter 11, Problem 28QAP

Diethylhydrazine reacts with iodine according to the following equation:

  ${\left({\text{C}}_2{\text{H}}_5\right)}_2{\left(\text{NH}\right)}_2\left(l\right)+{\text{I}}_2\left(aq\right)\to {\left({\text{C}}_2{\text{H}}_5\right)}_2{\text{N}}_2+2\text{HI}\left(aq\right)$The rate of the reaction is followed by monitoring the disappearance of the purple color due to iodine. The following data are obtained at a certain temperature.

(a) What is the order of the reaction with respect to diethylhydrazine, iodine, and overall?

(b) Write the rate expression of the reaction.

(c) Calculate k for the reaction.

(d) What must [(C₂H₅)₂] be so that the rate of the reaction is $5.00\times {10}^{-4}\text{mol}/\text{L}\cdot \text{h}$ when $\left[{\text{I}}_2\right]=0.500M$?

Interpretation Introduction

(a)

Interpretation:

To determine the order of reaction with respect to (C₂ H₅ )₂ (NH)₂, I₂ and overall for the following reaction:

${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{(NH)}}_{\text{2}}\text{(l)}\text{+}{\text{I}}_{\text{2}}\text{(aq)}\stackrel{}{\to }{{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}\text{(l)}\text{+}\text{2HI}\text{(aq)}$

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 28QAP

Order of given reaction:

With respect to (C₂ H₅ )₂ (NH)₂ =1

With respect to I₂ =1

Overall =2.

Explanation of Solution

Given information:

Here the chemical reaction is:

${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{(NH)}}_{\text{2}}\text{(l)}\text{+}{\text{I}}_{\text{2}}\text{(aq)}\stackrel{}{\to }{{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}\text{(l)}\text{+}\text{2HI}\text{(aq)}$

Let’s assume the reaction to be ‘t’ order with respect to (C₂ H₅ )₂ (NH)₂ and ‘y’ order with respect to I₂.

Then, rate law for experiment 1, 2, 3 and 4 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}\text{k[}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{2}}{\left(\text{NH}\right)}_{\text{2}}{\text{]}}^{\text{t}}{{\text{[I}}_{\text{2}}}^{\text{]}}\\ \text{1}{\text{.08×10}}^{\text{-4}}\text{=}\text{k[0}{\text{.150]}}^{\text{t}}{\text{[0}}^{\text{.250]}}\mathrm{......}\text{(1)}\\ \text{1}{\text{.56×10}}^{\text{-4}}\text{=}\text{k[0}{\text{.150]}}^{\text{t}}{\text{[0}}^{\text{.3620]}}\mathrm{......}\text{(2)}\\ \text{2}{\text{.30×10}}^{\text{-4}}\text{=}\text{k[0}{\text{.200]}}^{\text{t}}{\text{[0}}^{\text{.400]}}\text{.}\mathrm{.....}\text{(3)}\\ \text{3}{\text{.44×10}}^{\text{-4}}\text{=}\text{k[0}{\text{.300]}}^{\text{t}}{\text{[0}}^{\text{.400]}}\mathrm{......}\text{(4)}\end{array}$

Dividing (1) by (2) to get value of ‘y’.

$\begin{array}{l}\frac{\text{1}{\text{.08×10}}^{\text{-4}}}{\text{1}{\text{.56×10}}^{\text{-4}}}\text{=}{\left(\frac{\text{0}\text{.250}}{\text{0}\text{.3620}}\right)}^{\text{y}}\\ \Rightarrow \text{0}\text{.691}\text{=}{\left(\text{0}\text{.691}\right)}^{\text{y}}\\ \Rightarrow \text{y=1}\end{array}$

Thus, order with respect to I₂ is 1

Dividing (3) by (4) to get value of ‘t’.

$\begin{array}{l}\frac{2.30\times {\text{10}}^{\text{-4}}}{3.44\times {\text{10}}^{\text{-4}}}\text{=}{\left(\frac{\text{0}\text{.200}}{\text{0}\text{.300}}\right)}^{\text{t}}\\ \Rightarrow 0.66\text{=}{\left(\text{0}\text{.66}\right)}^{\text{t}}\\ \Rightarrow \text{t=}\text{1}\end{array}$

Thus, order with respect to (C₂ H₅ )₂ (NH)₂ is 1.

And the order of reaction will be:

$\Rightarrow \text{ t + y = 1+1 =}\text{2}$

Thus, overall order of reaction is 2.

Interpretation Introduction

(b)

Interpretation:

To write the rate expression for the given reaction.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 28QAP

Rate law expression for above reaction will be;

$\text{rate}\text{=}\text{k[}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{2}}{\left(\text{NH}\right)}_{\text{2}}{\text{][I}}_{\text{2}}\text{]}$

Explanation of Solution

Here the chemical reaction is:

${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{(NH)}}_{\text{2}}\text{(l)}\text{+}{\text{I}}_{\text{2}}\text{(aq)}\stackrel{}{\to }{{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}\text{(l)}\text{+}\text{2HI}\text{(aq)}$

Order of reaction with respect to (C₂ H₅ )₂ (NH)₂ = 1

Order of reaction with respect to I₂ = 1

Let the rate constant be ‘k’.

Then, rate law expression for above reaction will be;

$\text{rate}\text{=}\text{k[}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{2}}{\left(\text{NH}\right)}_{\text{2}}{\text{][I}}_{\text{2}}\text{]}$

Interpretation Introduction

(c)

Interpretation:

To determine the rate constant and its unit for the given reaction.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 28QAP

Rate constant is $2.88\times {10}^{-3}\text{L/mol}\text{.h}$

Explanation of Solution

Here the chemical reaction is:

${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{(NH)}}_{\text{2}}\text{(l)}\text{+}{\text{I}}_{\text{2}}\text{(aq)}\stackrel{}{\to }{{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}\text{(l)}\text{+}\text{2HI}\text{(aq)}$

Writing rate law for experiment 1 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}\text{k[}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{2}}{\left(\text{NH}\right)}_{\text{2}}{\text{][I}}_{\text{2}}\text{]}\\ \text{1}{\text{.08×10}}^{\text{-4}}\left(\frac{\text{mol}}{\text{L}\text{.h}}\right)\text{=}\text{k}\left(\text{U}\right)\text{×0}\text{.150}\left(\frac{\text{mol}}{\text{L}}\right)\text{×0}\text{.250}\left(\frac{\text{mol}}{\text{L}}\right)\\ \text{where}\text{U}\text{represent}\text{unit}\text{of}\text{rate}\text{constant}\text{.}\\ \Rightarrow \text{k}\left(\text{U}\right)\text{=}\frac{\text{1}{\text{.08×10}}^{\text{-4}}\left(\frac{\text{mol}}{\text{L}\text{.h}}\right)}{\text{0}\text{.150}\left(\frac{\text{mol}}{\text{L}}\right)\times 0.250{\left(\frac{\text{mol}}{\text{L}}\right)}^{}}=2.88\times {10}^{-3}\text{L/mol}\text{.h}\\ \end{array}$

Hence, the rate constant is $2.88\times {10}^{-3}\text{L/mol}\text{.h}$

Interpretation Introduction

(d)

Interpretation:

To determine the concentration of (C₂ H₅ )₂ (NH)₂.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 28QAP

Concentration of (C₂ H₅ )₂ (NH)₂ is 0.347 mol/L

Explanation of Solution

Here the chemical reaction is:

${{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{(NH)}}_{\text{2}}\text{(l)}\text{+}{\text{I}}_{\text{2}}\text{(aq)}\stackrel{}{\to }{{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{)}}_{\text{2}}{\text{N}}_{\text{2}}\text{(l)}\text{+}\text{2HI}\text{(aq)}$

Rate law expression for above reaction:

$\text{rate}\text{=}\text{k[}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\right)}_{\text{2}}{\left(\text{NH}\right)}_{\text{2}}{\text{][I}}_{\text{2}}\text{]}$

Here we have:

[(C₂ H₅ )₂ (NH)₂ ]= let it ‘y’ M

[I₂ ] = 0.500 M

Rate constant = $2.88\times {10}^{-3}\text{L/mol}\text{.h}$

Rate of reaction = 5.00×10-4 mol/L.h

Plugging values in rate law as:

$\begin{array}{l}\text{5}{\text{.00×10}}^{\text{-4}}\text{=}\text{2}{\text{.88×10}}^{\text{-3}}\text{×(y)×(0}{\text{.500)}}^{}\\ \Rightarrow \text{y}\text{=}\text{0}\text{.347}\text{mol/L}\end{array}$

Hence, the concentration of (C₂ H₅ )₂ (NH)₂ is 0.347 mol/L