C. The reactant concentration in a first-order reaction was 6.10×10 ?2 M after 3

Question

C. The reactant concentration in a first-order reaction was 6.10×10?2M after 30.0 s and 7.20×10?3M after 85.0s . What is the rate constant for this reaction?

D. The reactant concentration in a second-order reaction was 0.320 M after 240 s and 6.40×10?2Mafter 700 s . What is the rate constant for this reaction?

For part C the answer is NOT 4.330*10^-2 s^-1

For part D the answer is NOT 5.43*10^-3 mole^-1. lit .s^-1

Please show the work and the correct untis for each! thank you!

The integrated rate laws for zero-, first-, and second order reaction may be arranged such that they resemble the equation for a straight line, Order Integrated Rate LawGrap Slope [A] = _ kt + [A]0 | [A] vs. t | -k vs.t k

Explanation / Answer

C. The reactant concentration in a first-order reaction was 6.10×102M after 30.0 s and 7.20×103M after 85.0s . What is the rate constant for this reaction?

Solution :-

Change in concentration = 6.10*10^-2 – 7.20*10^-3 = 5.48*10^-2

Change in time = 85 s – 30.0 s = 55 s

If the initial concentration is 6.10*10^-2 M then half of it is (6.1*10^2/2)=3.05*10^-2 M

Now lets calculate the half life

3.05*10^-2 M * 55 sec / 5.48*10^-2 M =31.11 sec

Now using the half life lets calculate the rate constant K

K= 0.693 / t1/2

K= 0.693 / 31.11 s

K= 2.23*10^-2 s-1

Therefore rate constant = 2.23*10^-2 s-1

D. The reactant concentration in a second-order reaction was 0.320 M after 240 s and 6.40×102Mafter 700 s . What is the rate constant for this reaction?

Solution :-

Lets calculate the change in concentration

Change in concentration = 0.320 M – 0.0640 M = 0.256 M

Change in time = 700 s – 240 s = 460 s

Lets assume 0.320 is the initial concentration

Then half of it is 0.320 M/ 2 =0.160 M

Now lets calculate the half life

0.160 M * 460 s/0.256 M = 287.5 s

Now using the half life lets calculate the rate constant

K= (1/t1/2)[ A]o

K= (1/287.5 s)*0.230 M

K= 8.0*10^-4 M-1.s-1

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