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

1) The reactant concentration in a zero-order reaction was 6.00×10⁻²M after 150 sand 3.50×10⁻²M after 330 s . What is the rate constant for this reaction? Express your answer with the appropriate units.

2) What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units.

3) The reactant concentration in a first-order reaction was 9.20×10⁻² M after 50.0 s and 7.50×10⁻³ M after 60.0 s . What is the rate constant for this reaction? Express your answer with the appropriate units.

4) The reactant concentration in a second-order reaction was 0.440 M after 180 s and 5.70×10⁻² M after 860 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-meter as N*m.

Answer 1

We use given integrated rate laws.

1)

Initial concentration of reactant = 6.00 E-2 M , After 150 second , 3.50 E-2 M

After 330 s

We use initial concentration [A]0 = 6.00 E-2 M , and final concentration [A]t = 3.50 E-2

Here T = 330 s – 150 s = 180 s

[A]_t = -kt + [A]₀

k =- ([A]_t - [A]₀ ) / t

k = -(3.50E-2 – 6.0 E-2)/ 180 s

= 1.4 E-4 per s

Or 1.4 E-4 s^-1

2)

Here we find the concentration of reactant at time = 0 s

[A]_t = -kt + [A]₀

We use concentration of At at t = 150 s

[A]₀ = [A]_t +kt

= 6.00 E-2 M + (1.4 E-4 per s x 150 s )

= 8.1 E-2 M

3)

We use first order integrated rate equation.

Ln ([A]_t /[A]₀ =-kt

[A]₀ = 9.20 E-2 , t = 50.0 s

[A]_t = 7.50 E-2 M at 860 s    so time for decomposition = 860 – 50.0 = 630 s

Ln ([A]_t /[A]₀ =-kt

ln ( 7.50 E-2 / 9.20 E-2) = - k x 630 s

k = 3.24 E -4 S^-1

4)

[A]_t = 5.70 E-2 M   [A]₀ = 0.440 M  

t = 680 – 180 = 730 s

We use integrated second law for it.

1[A] = kt + [A]0

5.70 E-2 = k x 730 + 0.440

k = 4.7E-4 s^-1