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This reaction was monitored as a function of time: AB rightarrow A + B. A plot o

ID: 483276 • Letter: T

Question

This reaction was monitored as a function of time: AB rightarrow A + B. A plot of 1/[AB] versus time yields a straight line with a slope of +0.55/M s. a. What is the value of the rate constant (k) for this reaction at this temperature? b. Write the rate law for the reaction. c. What is the half-life when the initial concentration is 0.55 M? d. If the initial concentration of AB is 0.250 M, and the reaction mixture initially contains no products, what are the concentrations of A and B after 75 s? The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration. How long does it take for 25% of the C-14 atoms in a sample of C-14 to decay? If a sample of C-14 initially contains 1.5 mmol of C-14, how many mill moles are left after 2255 years?

Explanation / Answer

52. The plot of 1/[AB] vs time yields straight line, this means that the reaction is 2nd order.

a) Value of rate constant = slope of the line = 0.55 /M.s

b) rate law = k [AB]2

c) For 2nd order reaction, half life formula is: t1/2 = 1/k[AB]o

[AB]o = 0.55 M

k = 0.55 /M.s

So, t1/2 = 1/k[AB]o = 1/(0.55)(0.55) = 0.3025 s

d) Integrated rate law for 2nd order reaction: 1/[AB] = kt + 1/[AB]o

= (0.55)(75) + 1/0.250

= 41.25 + 4 = 45.25

[AB] = 1/45.25 = 0.022 M

[A] = 0.250-0.022 = 0.228 M

[B] = 0.228 M

56) Half life is independent of initial concentration for first order reaction.

t1/2 = 5730 yr

If initial concentration = 1 M

Final concentration = 1-(0.25)(1) = 0.75 M

rate constant = k = 0.693/t1/2 = 0.693/5730 = 0.00012 yr-1

for first order, ln [A] = ln [A]o - kt

ln 0.75 = ln 1 - 0.00012 t

-0.29 = 0 - 0.00012 t

t = 2416 yr

Time taken for 25 % decay = 2416 yr

ln [A] = ln [A]o - kt

= ln (1.5) - (0.00012) (2255)

= 0.405 - 0.271 = 0.134

ln [A] = 0.134

[A] = 1.143 mmol

[A] left after 2255 years = 1.143 mmol

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