The integrated rate laws for zero-, first-, and second-order reaction may be arr

Question

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.

k

Part A

The reactant concentration in a zero-order reaction was 8.00×102M after 200 s and 2.50×102Mafter 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 9.40×102M after 50.0 s and 1.30×103Mafter 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 7.70×102Mafter 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.

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

Explanation / Answer

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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