CHEM HELP!!! PLEASE HELP The integrated rate laws for zero-, first-, and second-

Question

CHEM HELP!!! PLEASE HELP

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.

Part A

The reactant concentration in a zero-order reaction was 7.00×102M after 140 s and 4.00×102M after 310 s . What is the rate constant for this reaction?

Part B

What was the initial reactant concentration for the reaction described in Part A?

Part C

The reactant concentration in a first-order reaction was 5.00×102M after 45.0 s and 1.00×102M after 100 s . What is the rate constant for this reaction?

Part D

The reactant concentration in a second-order reaction was 0.430 M after 260 s and 6.80×102M after 730 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-

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

a) For the zero order reaction

[Ao] = [At] + kt

7.00 * 10^(-2) = 4.00 * 10^(-2) + k[310-140]

3 * 10^(-2) = k(170)

k = 1.7647 * 10^(-4) Ms^(-1)

b) Initial concentration will be

[Ao] = [At] + kt

[Ao] = 7.00 * 10^(-2) + 1.7647 * 10^(-4) * 140

[Ao] = 7.00 * 10^(-2) + 2.47 * 10^(-2) = 9.47 * 10^(-2)

c)

For the first order reaction

ln(Ao/At) = kt

ln(5) = k(100-45)

k = ln(5)/55 = 2.92 * 10^(-2) s^(-1)

d)

1/At = 1/Ao + kt

1/(6.80*10^(-2)) = 1/0.430 + k(730-260)

12.3803 = k(470)

k = 12.3803/470 = 2.6341 * 10^(-2) M^(-1) s^(-1)

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