Question

The decomposition of N2O5 can be described by the equation.

2N2O5 (soln) --->  4NO2 (soln) + 2 (g)

Given this data for the reaction at 45 degrees C in carbon tetrachloride solution, calculate the average rate for each successive time interval.

t(s)                [N2O5] (M)

0                    2.10

195                1.86

556                1.48

825                1.25

Interval: 0 s to 195 s

Reaction rate= _____M/s

195 s to 556 s

Reaction rate= _____M/s

556 s to 825 s

Reaction rate= _____M/s

Answer 1

General guidance

Concepts and reason

This is based upon the concept of the rate of reaction which is the speed of the reaction or tells us how quickly or slowly a reaction will take place. It also tells us how quickly or slowly a reactant will disappear or a product will be generated.

Fundamentals

Average rate of a reaction is expressed in terms of reactant and product as follows:

Consider a reaction:

${\\rm{aP + bQ}} \\to {\\rm{xM + yN}}$

Here, a, b, x and y are the stoichiometric coefficients, while P and Q are the reactants and M and N are the products of the reaction.

As we know, rate of reaction (r) is defined as the rate of disappearance of reactant or rate of production of products. So, mathematically, it is represented as

$r = {\\rm{ - }}\\frac{{\\rm{1}}}{{\\rm{a}}}\\frac{{{\\rm{d}}\\left[ {\\rm{P}} \\right]}}{{{\\rm{dt}}}}{\\rm{ = - }}\\frac{{\\rm{1}}}{{\\rm{b}}}\\frac{{{\\rm{d}}\\left[ {\\rm{Q}} \\right]}}{{{\\rm{dt}}}}{\\rm{ = }}\\frac{{\\rm{1}}}{{\\rm{x}}}\\frac{{{\\rm{d}}\\left[ {\\rm{M}} \\right]}}{{{\\rm{dt}}}}{\\rm{ = }}\\frac{{\\rm{1}}}{{\\rm{y}}}\\frac{{{\\rm{d}}\\left[ {\\rm{N}} \\right]}}{{{\\rm{dt}}}}$

Step-by-step

Step 1 of 4

The provided reaction is

Expression for average rate of disappearance of ${{\\rm{N}}_2}{{\\rm{O}}_5}$, r is as follows

$r = {\\rm{ - }}\\frac{{\\rm{1}}}{2}\\frac{{{\\rm{d}}\\left[ {{{\\rm{N}}_2}{{\\rm{O}}_5}} \\right]}}{{{\\rm{dt}}}}$

Here, ${\\rm{d}}\\left[ {{{\\rm{N}}_2}{{\\rm{O}}_5}} \\right]$ is change of concentration of ${{\\rm{N}}_2}{{\\rm{O}}_5}$and ${\\rm{dt}}$ it is the change in time.

Explanation | Common mistakes | Hint for next step

As the reaction proceeds, the concentration of reactants decreases. And the average rate of decrease in concentration of reactant is the average rate of the reaction.

Step 2 of 4

For the first case,

$\\begin{array}{c}\\\\{\\rm{d}}\\left[ {{{\\rm{N}}_2}{{\\rm{O}}_5}} \\right] = 1.86{\\rm{ M}} - 2.1{\\rm{ M}}\\\\\\\\ = - 0.24{\\rm{ M}}\\\\\\end{array}$

And

$\\begin{array}{c}\\\\{\\rm{dt}} = {\\rm{195 s}} - {\\rm{0 s}}\\\\\\\\{\\rm{ = 195 s}}\\\\\\end{array}$

Substitute $- 0.24{\\rm{ M}}$for ${\\rm{d}}\\left[ {{{\\rm{N}}_2}{{\\rm{O}}_5}} \\right]$and 195 s for dt

$\\begin{array}{c}\\\\r = - \\frac{{\\rm{1}}}{2}\\frac{{\\left( { - 0.24{\\rm{ M}}} \\right)}}{{\\left( {{\\rm{195 s}}} \\right)}}\\\\\\\\ = 0.0006154{\\rm{ M}}{{\\rm{s}}^{ - 1}}\\\\\\\\ = 6.1 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}\\\\\\end{array}$

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 0 s to 195 s is $6.1 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$

As the reaction proceeds, the concentration of reactants decreases. And the average rate of decrease in concentration of reactant is the average rate of the reaction.

Step 3 of 4

[Step 3

For the second case,

$\\begin{array}{c}\\\\{\\rm{d}}\\left[ {{{\\rm{N}}_2}{{\\rm{O}}_5}} \\right] = 1.48{\\rm{ M}} - 1.86{\\rm{ M}}\\\\\\\\ = - 0.38{\\rm{ M}}\\\\\\end{array}$

And

$\\begin{array}{c}\\\\{\\rm{dt}} = 556{\\rm{ s}} - {\\rm{195 s}}\\\\\\\\{\\rm{ = 361 s}}\\\\\\end{array}$

Substitute $- 0.38{\\rm{ M}}$for ${\\rm{d}}\\left[ {{{\\rm{N}}_2}{{\\rm{O}}_5}} \\right]$and 361 s for dt

$\\begin{array}{c}\\\\r = - \\frac{{\\rm{1}}}{2}\\frac{{\\left( { - 0.38{\\rm{ M}}} \\right)}}{{\\left( {{\\rm{361s}}} \\right)}}\\\\\\\\ = 0.0005263{\\rm{ M}}{{\\rm{s}}^{ - 1}}\\\\\\\\ = 5.2 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}\\\\\\end{array}$

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 195 s to 556 s is $5.2 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$

As the reaction proceeds, the concentration of reactants decreases. And the average rate of decrease in concentration of reactant is the average rate of the reaction.

Step 4 of 4

For the third case,

$\\begin{array}{c}\\\\{\\rm{d}}\\left[ {{{\\rm{N}}_2}{{\\rm{O}}_5}} \\right] = 1.25{\\rm{ M}} - 1.48{\\rm{ M}}\\\\\\\\ = - 0.23{\\rm{ M}}\\\\\\end{array}$

And

$\\begin{array}{c}\\\\{\\rm{dt}} = {\\rm{825 s}} - {\\rm{556 s}}\\\\\\\\{\\rm{ = 269 s}}\\\\\\end{array}$

Substitute $- 0.23{\\rm{ M}}$for ${\\rm{d}}\\left[ {{{\\rm{N}}_2}{{\\rm{O}}_5}} \\right]$and 269 s for dt

$\\begin{array}{c}\\\\r = - \\frac{{\\rm{1}}}{2}\\frac{{\\left( { - 0.23{\\rm{ M}}} \\right)}}{{\\left( {{\\rm{269 s}}} \\right)}}\\\\\\\\ = 0.0004275{\\rm{ M}}{{\\rm{s}}^{ - 1}}\\\\\\\\ = 4.2 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}\\\\\\end{array}$

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 825 s to 556 s is $4.2 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$

As the reaction proceeds, the concentration of reactants decreases. And the average rate of decrease in concentration of reactant is the average rate of the reaction.

Answer

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 0 s to 195 s is $6.1 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 195 s to 556 s is $5.2 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 825 s to 556 s is $4.2 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$

Answer only

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 0 s to 195 s is $6.1 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 195 s to 556 s is $5.2 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$

The average rate of change of ${{\\rm{N}}_2}{{\\rm{O}}_5}$ from 825 s to 556 s is $4.2 \\times {10^{ - 4}}{\\rm{ M}}{{\\rm{s}}^{ - 1}}$