Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically, but it can also be calculated using calculus. The reaction of compound A forming compound B was studied and the following data were collected:

Time (s) [A] (M)
0. 0.184
200. 0.129
500. 0.069
800. 0.031
1200. 0.019
1500. 0.016

Required:
a. What is the average reaction rate between 0 and 1500s?
b. What is the average reaction rate between 200. and 1200s ?
c. What is the instantaneous rate of the reaction at t=800s?

Sagot :

Answer:

Explanation:

a )

Average reaction rate between 0 and 1500s

Time duration = 1500 s

moles reacted = .184 - .016 = .168 moles

Moles reacted per second = .168 / 1500

= 112 x 10⁻⁶ moles /s

b )

Average reaction rate between 200 and 1200s

Time duration = 1000 s

moles reacted = .129 - .019 = .11 moles

Moles reacted per second = .11 / 1000

= 110 x 10⁻⁶ moles /s

c )

the instantaneous rate of the reaction at t=800s

We shall assume that between 500 s and 1200 s , rate of reaction is uniform

rate between 500 and 1200

Time duration = 700 s

moles consumed = .069 - .019 = .05 moles

Rate of reaction = .05 / 700

= 71 .4 x 10⁻⁶ moles / s

This will also be instantaneous rate of reaction at t = 800 s .

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.