Question

The reaction

A+ B yields C+D   

rate = k[A][B]^2

has an intial rate of 0.0220 M/s

what will the initial rate be if [A] is halved and [B] is tripled? answer must be in M/s

What will the intial rate be if [A] is tripled and [B] is halved?  answer must be in M/s

Answer 1

General guidance

Concepts and reason

The concept used to solve this problem is based on rate law of a chemical reaction.

Rate law is an equation that establishes a link between the concentration of the reactants and the rate of the reaction.

Fundamentals

For a chemical reaction,

${\\rm{A}} + {\\rm{B}} \\to {\\rm{C}}$

The rate law is written as follow.

$r = k{\\left[ {\\rm{A}} \\right]^x}{\\left[ {\\rm{B}} \\right]^y}$

Here, $\\left[ {\\rm{A}} \\right]$ is the concentration of reactant ${\\rm{A}}$, $\\left[ {\\rm{B}} \\right]$ is the concentration of reactant ${\\rm{B}}$,$k$ is the rate constant of the reaction,$x$ and $y$ are the reaction orders which is an experimentally determined quantity.

Step-by-step

Step 1 of 3

Part 1

The reaction is:

The rate law for this reaction is written as:

$r = k\\left[ {\\rm{A}} \\right]{\\left[ {\\rm{B}} \\right]^2}$

Substitute initial rate of the reaction $\\left( r \\right)$ $0.0220{\\rm{ m }}{{\\rm{s}}^{ - 1}}$ in rate law.

$0.0220{\\rm{ M/s}} = k\\left[ {\\rm{A}} \\right]{\\left[ {\\rm{B}} \\right]^2}$

Explanation | Hint for next step

An expression for initial rate is obtained using the rate law by substituting the value of initial rate.

Step 2 of 3

Part 1

When $\\left[ {\\rm{A}} \\right]$ is halved and $\\left[ {\\rm{B}} \\right]$ is tripled, the new concentrations are:

$\\left[ {{\\rm{A'}}} \\right] = \\frac{{\\left[ {\\rm{A}} \\right]}}{{\\rm{2}}}$ and $\\left[ {{\\rm{B'}}} \\right] = \\left[ {\\rm{B}} \\right] \\times 3$

The new rate law is written as:

$r = k\\left[ {{\\rm{A'}}} \\right]{\\left[ {{\\rm{B'}}} \\right]^2}$

Substitute, $\\frac{{\\left[ {\\rm{A}} \\right]}}{{\\rm{2}}}$ for $\\left[ {{\\rm{A'}}} \\right]$ and $3\\left[ {\\rm{B}} \\right]$ for $\\left[ {{\\rm{B'}}} \\right]$ in new rate law as follows:

$\\begin{array}{c}\\\\r = k\\frac{{\\left[ {\\rm{A}} \\right]}}{{\\rm{2}}}{\\left( {3\\left[ {\\rm{B}} \\right]} \\right)^2}\\\\\\\\ = \\frac{9}{2}k\\left[ {\\rm{A}} \\right]{\\left[ {\\rm{B}} \\right]^2}\\\\\\end{array}$ \u2026\u2026 (1)

Since,

$k\\left[ {\\rm{A}} \\right]{\\left[ {\\rm{B}} \\right]^2} = 0.0220{\\rm{ M }}{{\\rm{s}}^{ - 1}}$

Substitute, $0.0220{\\rm{ m }}{{\\rm{s}}^{ - 1}}$ for $k\\left[ {\\rm{A}} \\right]{\\left[ {\\rm{B}} \\right]^2}$ in equation (1), thus

$\\begin{array}{c}\\\\r = \\frac{9}{2} \\times 0.0220{\\rm{ m }}{{\\rm{s}}^{ - 1}}\\\\\\\\ = {\\rm{0}}{\\rm{.099 m }}{{\\rm{s}}^{ - 1}}\\\\\\end{array}$

Rate of the reaction become ${\\rm{0}}{\\rm{.099 m }}{{\\rm{s}}^{ - 1}}$.

Part 1

The rate of the reaction when $\\left[ {\\rm{A}} \\right]$ is halved and $\\left[ {\\rm{B}} \\right]$ is tripled is ${\\rm{0}}{\\rm{.099 m }}{{\\rm{s}}^{ - 1}}$.

Explanation | Common mistakes | Hint for next step

When the initial concentrations of the reactants are changed by any factor, then the first step is to write the new concentrations in terms of the initial concentrations. Then, write the new rate law in terms of the initial concentrations. And then calculate the new rate law by substituting the values accordingly.

Step 3 of 3

Part 2

When $\\left[ {\\rm{A}} \\right]$ is tripled and $\\left[ {\\rm{B}} \\right]$ is halved, so the new concentrations are:

$\\left[ {{\\rm{A''}}} \\right] = 3 \\times \\left[ {\\rm{A}} \\right]$ and $\\left[ {{\\rm{B''}}} \\right] = \\frac{{\\left[ {\\rm{B}} \\right]}}{2}$

The new rate law is written as:

$r = k\\left[ {{\\rm{A''}}} \\right]{\\left[ {{\\rm{B''}}} \\right]^2}$

Substitute, $3\\left[ {\\rm{A}} \\right]$ for $\\left[ {{\\rm{A'}}} \\right]$ and $\\frac{{\\left[ {\\rm{B}} \\right]}}{2}$ for $\\left[ {{\\rm{B'}}} \\right]$ in new rate law as follows:

$\\begin{array}{c}\\\\r = k\\left( {3 \\times \\left[ {\\rm{A}} \\right]} \\right){\\left( {\\frac{{\\left[ {\\rm{B}} \\right]}}{2}} \\right)^2}\\\\\\\\ = \\frac{3}{4}k\\left[ {\\rm{A}} \\right]{\\left[ {\\rm{B}} \\right]^2}\\\\\\end{array}$ \u2026\u2026 (2)

Since,

$k\\left[ {\\rm{A}} \\right]{\\left[ {\\rm{B}} \\right]^2} = 0.0220{\\rm{ M }}{{\\rm{s}}^{ - 1}}$

Substitute, $0.0220{\\rm{ m }}{{\\rm{s}}^{ - 1}}$ for $k\\left[ {\\rm{A}} \\right]{\\left[ {\\rm{B}} \\right]^2}$ in equation (2), thus

$\\begin{array}{c}\\\\r = \\frac{3}{4} \\times 0.0220{\\rm{ m }}{{\\rm{s}}^{ - 1}}\\\\\\\\ = {\\rm{0}}{\\rm{.0165 m }}{{\\rm{s}}^{ - 1}}\\\\\\end{array}$

Rate of the reaction become ${\\rm{0}}{\\rm{.0165 m }}{{\\rm{s}}^{ - 1}}$.

Part 2

The rate of the reaction when $\\left[ {\\rm{A}} \\right]$ is tripled and $\\left[ {\\rm{B}} \\right]$ is halved is ${\\rm{0}}{\\rm{.0165 m }}{{\\rm{s}}^{ - 1}}$.

When the initial concentrations of the reactants are changed by any factor, then the first step is to write the new concentrations in terms of the initial concentrations. Then, write the new rate law in terms of the initial concentrations. And then calculate the new rate law by substituting the values accordingly.

Answer

Part 1

The rate of the reaction when $\\left[ {\\rm{A}} \\right]$ is halved and $\\left[ {\\rm{B}} \\right]$ is tripled is ${\\rm{0}}{\\rm{.099 m }}{{\\rm{s}}^{ - 1}}$.

Part 2

The rate of the reaction when $\\left[ {\\rm{A}} \\right]$ is tripled and $\\left[ {\\rm{B}} \\right]$ is halved is ${\\rm{0}}{\\rm{.0165 m }}{{\\rm{s}}^{ - 1}}$.

Answer only

Part 1

The rate of the reaction when $\\left[ {\\rm{A}} \\right]$ is halved and $\\left[ {\\rm{B}} \\right]$ is tripled is ${\\rm{0}}{\\rm{.099 m }}{{\\rm{s}}^{ - 1}}$.

Part 2

The rate of the reaction when $\\left[ {\\rm{A}} \\right]$ is tripled and $\\left[ {\\rm{B}} \\right]$ is halved is ${\\rm{0}}{\\rm{.0165 m }}{{\\rm{s}}^{ - 1}}$.