Question

PLEASE MAKE SURE YOUR ANSWERS ARE CORRECT

Part A

The reactant concentration in a zero-order reaction was 0.100 M after 105 s and 3.00×10−2 M after 350 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 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 7.50×10⁻²M after 45.0 s and 7.60×10⁻³M after 100 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

Part D

The reactant concentration in a second-order reaction was 0.590 M after 120 s and 4.90×10⁻²M after 755 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

for a zero order reaction, the equation is

CA= CAO-Kt, CAO= initial concentration, CA= Concentration at any time, t and K= rate constant and t is the time

For the given conditions is

0.1= CAO- 105K (1)

3/100= CAO- 350K (2)

Eq.1- Eq.2 = 0.1-0.3/100 = (350-105)K

K= 0.000286 M/sec

From Eq.1, 0.1= CAO-105*0.000286

CAO= 0.13M

3.

For a 1^st order reaction

lnCA= lnCAO- Kt

CAO= initial concentration, CA= Concentration at any time, t and K= rate constant and t is the time

from the given information

ln(7.5*10^-2)= lnCAO-45K (1)

ln(7.6*10^-3)= lnCAO- 100K (2)

Eq.1-Eq.2 ln (7.5*10^-2/ 7.6*10^-3)= (100-45)K

K=0.042/sec

4. for second order reaction 1/CA= 1/CAO+Kt,CAO= initial concentration, CA= Concentration at any time, t and K= rate constant

hence 1/0.59= 1/CAO+K*120, 1.69= 1/CAO+K*120 (1) and 1/(4.9*10^-2)= 1/CAO+K*755, 20.41= 1/CAO+755K (2)

Eq.2-Eq.1= 20.41-1.69= K*(755-120), K= 0.02948/M.sec