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.

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

a.)

The reactant concentration in a zero-order reaction was 6.00×10⁻²M after 140 s and 2.00×10⁻²Mafter 310 s . What is the rate constant for this reaction?

b.)

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

c.)The reactant concentration in a first-order reaction was 6.20×10⁻²M after 15.0 s and 1.00×10⁻³Mafter 85.0 s . What is the rate constant for this reaction?

d.)The reactant concentration in a second-order reaction was 0.560 M after 155 s and 5.40×10⁻²M after 740 s . What is the rate constant for this reaction?

Answer 1

(a).

For a zero order reaction

[A] = -kt + [A°]

[A°] = kt + [A]

At t = 140s [A] = 6.00*10^-2M

So,

[A°] = k*140s + 6.00*10^-2M

At t = 310s [A] = 2.00*10^-2M

[A°] = k*310s + 2.00*10^-2M

Equating both equations,

k*140s + 6.00*10^-2M = k*310s + 2.00*10^-2M

6.00*10^-2M - 2.00*10^-2M = 310s*k - 140s*k

4.00*10^-2M = 170s*k

k = 4.00*10^-2M / 170s

k = 0.02*10^-2 M/s

(b) at t = 140s [A] = 6.00*10^-2M k = 0.02*10^-2M/s

[A]= -kt + [A°]

[A°] = kt + [A]

= 0.02*10^-2M/s*140s + 6.00*10^-2M

= 3.29*10^-2M + 6.00*10^-2M

[A°] = 9.29*10^-2M

Part C :-

For a first order reaction-

ln[A] = -kt + ln[A°]

ln[A°] = kt + ln[A]

When t = 15s [A] = 6.20*10^-2M

So,

ln[A°] = k*15s + ln(6.20*10^-2M)

At t = 85s [A]= 1.00*10^-3M

ln[A°] = 85s*k + ln(1.00*10^-3M)

Equating both-

k*15s + ln(6.20*10^-2M) = k*85s + ln(1.00*10^-3M)

ln(6.20*10^-2M) - ln(1.00*10^-3M) = k*85s - k*15s

ln(6.20*10^-2M/1.00*10^-3M) = 70s*k

ln(62) = 70s*k

4.127 = 70s*k

k = 0.059 s^-1

Part d:

For second order reaction-

1/[A] = 1/[A°] + kt

1/ [A°] = 1/[A] - kt

At t = 155s [A] = 0.560M

So,

1/ [A° ] = 1/0.560M - k*155s

At t = 740s [A] = 5.40*10^-2M

1/ [A°] = 1/ 5.40*10^-2M - k*740s

Equating both,

1/ 0.560M - k*155s = 1/ 5.40*10^-2M - k*740s

-k*155s + k*740s = 1/5.40*10^-2M - 1/0.560M

585sk = 18.5185M^-1 - 1.7857M^-1

585sk = 16.733M^-1

k = 16.733M^-1/585s = 0.029M^-1/s

k = 0.029M^-1/s