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.

1) The reactant concentration in a zero-order reaction was 6.00×102M after 150 s and 3.50×102Mafter 330 s . What is the rate constant for this reaction? Give appropriate units.

2) The reactant concentration in a first-order reaction was 9.20×102M after 50.0 s and 7.50×103Mafter 60.0 s . What is the rate constant for this reaction? Give appropriate units.

3)The reactant concentration in a second-order reaction was 0.440 M after 180 s and 5.70×102 Mafter 860 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

1) For the first reaction which is zero order reaction

[Ao] = [At] + kt

6.00 * 10^(-2) = 3.50 * 10^(-2) + k(180)

k = 1.388* 10^(-4) M L^(-1) s^(-1)

2) For the first reaction which is first order reaction

ln(Ao/At) = kt

ln(9.20 * 10^(-2)/7.50 * 10^(-3)) = k(60-50)

k = 0.25068 s^(-1)

3) For the third reaction which is second order reaction

1/[At] = 1/[A0] + kt

1/(5.70*10^(-2)) = 1/0.440 + k(860-180)

k = 0.022457 M^(-1) L s^(-1)

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