Does any one know how to answer the following rate question?

A certain reaction has the following general form.

aA bB

At a particular temperature and [A]0 = 2.00 10-2 M, concentration versus time data were collected for this reaction, and a plot of ln[A] versus time resulted in a straight line with a slope value of -2.97 10-2 min-1.

(a) Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction.

rate law

a. k[A]

b. k[A]2

c. k[A]3

d. k[B]

(b) integrated rate law

a. ln[A] = -kt - ln[A]0

b. ln[A] = -kt + ln(1/[A]0)

c. ln[A] = -kt + ln[A]0

d. ln[A] = kt - ln[A]0

(c) rate constant in min-1

.

(d) Calculate the half-life for this reaction in min

(e) How much time is required for the concentration of A to decrease to 2.00 10-3 M?

in min

Does any one know how to answer the following rate question?

(a)

ln[A] is proportional to t: So you can relate [A] and t as:

ln[A] = k璺痶 + m

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[A] = e^(k璺痶 + m)

(where k and m are arbitrary constants:

The rate law describes

-r = d[A]/dt

To find just take the derivative of equation above:

-r = d/dt{e^(k璺痶 + m)}

= k璺痚^(k璺痶 + m)

= k璺痆A]

answer a. is the correct one

(b)

d[A]/dt = -k璺痆A]

separate variables and integrate

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1/[A] d[A] = - k dt

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閳?1/[A] d[A] = - 閳?k dt

%26lt;=%26gt;

ln[A] = - k璺痶 + c

apply initial condition to find constant c

Apply initial condition

t = 0 閳?[A]=[A]閳р偓

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ln[A]閳р偓 = - k璺? + c

%26lt;=%26gt;

c = ln[A]閳р偓

Hence:

ln[A] = - k璺痶 + ln[A]閳р偓

answer c. is correct

(c)

From the equation derived in(b) you see that the slope of the line representing ln[A] vs t is equal to - k. Hence:

k = 2.97鑴?0^-2min-1

(d)

Half life is the time elapsed till [A] has dropped to one half of its initial value:

Therefore

ln( [A]閳р偓/2) = - k璺痶1/2 + ln[A]閳р偓

%26lt;=%26gt;

t1/2 = ( ln[A]閳р偓 - ln( [A]閳р偓/2) ) / k

= ln{ [A]閳р偓/ ([A]閳р偓/2) } / k

= ln{2) / k

= ln{2) / 2.97鑴?0^-2min-1

= 23.34min

(e)

ln[A] = - k璺痶 + ln[A]閳р偓

%26lt;=%26gt;

k璺痶 = ln[A]閳р偓 - ln[A] k

%26lt;=%26gt;

t = ln( [A]閳р偓/ [A] ) / k

= ln( 2.0鑴?0^-2M/ 2.0鑴?0^-3M ) / 2.97鑴?0^-2min-1

= 77.53min