Respuesta :
The rate equation is r = k[a]²[b]².
Take logarithms of both sides.
ln r = ln k + 2 ln [a] + 2 ln [b]
When [a] = 0.0.100 m, r₁ = 0.0200 m/s.
Therefore
ln(0.0200) = ln k + 2 ln (0.0100) + 2 ln [b] (1)
When a = 0.200 m, obtain the new rate, r₂, as
ln r₂ = ln k + 2 ln (0.0200) + 2 ln [b] (2)
Subtract (1) from (2) to obtain
ln r₂ - ln(0.0200) = 2{ln(0.0200 - ln(0.0100)}
ln(r₂/0.0200) = 2 ln(0.0200/0.0100)
= 2 ln(2)
= ln(2²) = ln(4)
Therefore
r₂/0.0200 = 4
r₂ = 0.0800 m/s
Answer: 0.0800 m/s
Take logarithms of both sides.
ln r = ln k + 2 ln [a] + 2 ln [b]
When [a] = 0.0.100 m, r₁ = 0.0200 m/s.
Therefore
ln(0.0200) = ln k + 2 ln (0.0100) + 2 ln [b] (1)
When a = 0.200 m, obtain the new rate, r₂, as
ln r₂ = ln k + 2 ln (0.0200) + 2 ln [b] (2)
Subtract (1) from (2) to obtain
ln r₂ - ln(0.0200) = 2{ln(0.0200 - ln(0.0100)}
ln(r₂/0.0200) = 2 ln(0.0200/0.0100)
= 2 ln(2)
= ln(2²) = ln(4)
Therefore
r₂/0.0200 = 4
r₂ = 0.0800 m/s
Answer: 0.0800 m/s
The new rate when the concentration of [A] is increased to 0.200 M is 0.0800 M/s.
Further explanation:
The rate of reaction is defined decrease in concentrations of reaction in unit time or increase in products concentration in unit time. Rate of reaction is determined by the rate law. The rate law is a relationship between rate constant and reactants of the reaction.
Consider a reaction,
[tex]a{\text{A}} + b{\text{B}}\to{\text{Products}}[/tex]
The rate of the above equation is expressed as,
[tex]{\text{rate}} = k{\left[{\text{A}}\right]^a}{\left[{\text{B}}\right]^b}[/tex] …… (1)
Here,
rate is the rate of the reaction.
k is the rate constant.
[tex]\left[{\text{A}}\right][/tex] and [tex]\left[{\text{B}}\right][/tex] are the concentration of reactant species.
a and b are the order of reaction for A and B.
The given rate law is,
[tex]{\text{rate}} = k{\left[{\text{A}}\right]^2}{\left[{\text{B}}\right]^2}[/tex]
When the initial
Substitute 0.100 M for [tex]\left[{\text{A}}\right][/tex] ,0.0200 M/s for rate in equation (1).
[tex]\begin{gathered}{\text{rate}} = k{\left[{\text{A}}\right]^2}{\left[{\text{B}}\right]^2}\hfill\\0.0200{\text{ M/s}} = k{\left( {0.100{\text{ M}}}\right)^2}{\left[{\text{B}}\right]^2}\hfill\\\end{gathered}[/tex]
Therefore, the initial rate law is,
[tex]0.0200{\text{M/s}} = k{\left( {0.100{\text{ M}}}\right)^2}{\left[{\text{B}}\right]^2}[/tex] …… (2)
For final rate law when the concentration of A is increased to 0.200 M.
Substitute 0.200 M for [tex]\left[{\text{A}}\right][/tex] in equation (1).
[tex]\begin{gathered}{\text{rate}} = k{\left[{\text{A}}\right]^2}{\left[{\text{B}}\right]^2}\hfill\\0.0200{\text{ M/s}} = k{\left( {0.100{\text{ M}}}\right)^2}{\left[{\text{B}}\right]^2}\hfill\\\end{gathered}[/tex]
Therefore, the final rate law is,
[tex]{\text{rat}}{{\text{e}}_2} = k{\left({0.200{\text{ M}}}\right)^2}{\left[{\text{B}}\right]^2}[/tex] …… (3)
To calculate the value of the final rate, divide equation (3) from equation (2).
[tex]\frac{{{\text{rat}}{{\text{e}}_2}}}{{0.0200{\text{ M/s}}}} = \frac{k}{k}\frac{{{{\left({0.200{\text{ M}}}\right)}^2}}}{{{{\left({0.100{\text{ M}}}\right)}^2}}}\times\frac{{{{\left[{\text{B}}\right]}^2}}}{{{{\left[{\text{B}}\right]}^2}}}[/tex]
After simplifying this equation we get,
[tex]\begin{aligned}{\text{rat}}{{\text{e}}_2} &= \frac{{{{\left({0.200{\text{ M}}}\right)}^2}}}{{{{\left({0.100{\text{ M}}}\right)}^2}}}\times 0.0200{\text{ M/s}}\\ &= {\left({\text{2}}\right)^2}\times{\text{0}}{\text{.0200 M/s}}\\ &= 0.0800{\text{ M/s}}\\\end{aligned}[/tex]
Thus the value of the final rate is 0.0800 M/s.
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Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Kinetics
Keywords: Rate of reaction, new rate of reaction, concentration, concentration of A increased to 2.00 M, initial rate, final rate, rate law.
