The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b.
Order | Integrated Rate Law | Graph | Slope |
0 | [A]=−kt+[A]0 | [A] vs. t | −k |
1 | ln[A]=−kt+ln[A]0 | ln[A] vs. t | −k |
2 | 1[A]= kt+1[A]0 | 1[A] vs. t |
k |
Part A
The reactant concentration in a zero-order reaction was 8.00×10^{−2}M after 200 s and 2.50×10^{−2}Mafter 390 s . What is the rate constant for this reaction?
Express your answer with the appropriate units.
Part B
What was the initial reactant concentration for the reaction described in Part A?
Express your answer with the appropriate units.
Part C
The reactant concentration in a first-order reaction was 9.40×10^{−2}M after 50.0 s and 1.30×10^{−3}Mafter 65.0 s . What is the rate constant for this reaction?
Express your answer with the appropriate units.
Part D
The reactant concentration in a second-order reaction was 0.880 M after 160 s and 7.70×10^{−2}Mafter 885 s . What is the rate constant for this reaction?
Express your answer with the appropriate units. Include an asterisk to indicate a compound unit with mulitplication, for example write a Newton-meter as N*m.
Part A
For zero order reaction
[Ao] = [At] + kt
8.00 * 10^(-2) = 2.50 * 10^(-2) + k(390-200)
k = 5.50 * 10^(-2)/190 = 2.8947 * 10^(-4) Ms^(-1)
Part B
[Ao] = [At] + kt
[Ao] = 8.00 * 10^(-2) + 2.8947 * 10^(-4) * 200
[Ao] = 8.00 * 10^(-2) + 5.789 * 10^(-2) = 13.789 * 10^(-2) M
Part C
ln(Ao/At) = kt
ln(9.40 * 10^(-2)/1.30 * 10^(-3)) = k(65-50)
ln(72.307) = 15k
k = 0.2853 s^(-1)
Part D
1/At = 1/Ao + kt
1/7.70 * 10^(-2) = 1/0.880 + k(885-160)
k = 0.0163 M^(-1) s^(-1)
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