Average rate of production from rate constant

Question

The hydrolysis of table sugar (sucrose) occurs by the following overall reaction :

sucrose + water → glucose + fructose and is first order in sucrose concentration with a rate constant of

3.50 × 10⁻³ min⁻¹. If a 0.500 M solution of sucrose is allowed to react for 245 min, what will be the average rate of production of glucose over that period of time in units of M/min?

The only thing I can really figure out is that at some point I'll probably need to use ln?

Answer 1

We need to use the integrated rate law expression for this problem ln [sucrose]_t/[sucrose]₀ = -kt. This becomes ln[sucrose]_t - ln[0.5] = -3.50 x 10^-3 x 245 min.

Solving for ln[sucrose]_t = -3.5 x 10^-3 x 245 +ln[0.5] = -1.55 taking the antilog of both sides.

[sucrose]_t = 0.212 M at t = 245 min.

As glucose is produced in a 1:1 mole ratio from sucrose, The concentration of glucose at time t will be 0.5 - 0.212 = 0.288 M. The average rate of glucose production will be 0.288 M / 245 min = 0.00118 M/min.

Answer 2

Here we go .....

first-order rate equation = k * [A]^1

reads as rate constant (k) times the concentration raises to the first power

rate = k * [A]

the integrated rate law for a first-order reaction is

,

where [A] is the concentration at time t and [A]₀ is the initial concentration at zero time (the start)

The first-order rate law is confirmed if  is in fact a linear function of time

Notice that the slope of the linear line is negative and is the rate constant

the given information:

k = 3.50 × 10⁻³ min⁻¹

t = 245 min

[A]₀ = 0.500 M

use a calculator to determine -kt and write it down

-kt = __________

use a calculator to take the natural log of [A]₀ and write it down

ln [A]₀ = __________

plug them and the time into the equation and it'll give you ln[A]; don't worry, you should get a negative value

now you'll need [A] ... it's what is left after 245 min

Do you know how to "inverse" that natural log on your calculator?

[A] = __________

Then, the difference between [A] and [A]₀ went to glucose, which is what you are asked

that concentration divided by time should given you the "average rate of production of glucose"

It's the average because it reads like "an amount of glucose is produced per min"

Hope this helps. Continue comment with questions for clarification. :)