Question

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

Part A

The reactant concentration in a zero-order reaction was 5.00×10⁻²M after 115 s and 3.00×10⁻²M after 325 s. What is the rate constant for this reaction?     K_0th = 9.52x10^-5 M/s

Part B

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

Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

Part C

The reactant concentration in a first-order reaction was 9.00×10^-2M after 45.0 s and 8.60×10⁻³M after 95.0 s. What is the rate constant for this reaction?

Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

Part D

The reactant concentration in a second-order reaction was 0.650 M after 100 s and 7.80×10⁻²M after 775 s. What is the rate constant for this reaction?

Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

Answer 1

part-A

K    = 1/t [ [A0]-[A] ]

       = 1/(325-115) (0.05-0.03)

      = 0.00476* (0.05-0.03)   = 9.52*10^-5 M/sec

part -B

initial concntration = [A0]   = 5*10^-2 M

part-C

K    = 2.303/t log[A0]/[A]

      = 2.303/(95-45) log0.09/0.0086

    = 2.303*1.0197/50   = 0.0469 sec^-1

part-D

K   = 1/t [1/[A]-1/[A0] ]

    = 1/(775-100) [ 1/0.078 -1/0.65]

    = 1/675 (12.82-1.54)

     = 0.00148*11.28    = 0.0167M^-1 sec^-1