What would the concentration of the solution be after 2 days?

Question

cometailcane · Answer

Time (min)	Concentration (M)
0	                 0.0927
8		         0.0825
19		         0.0686
30.		         0.0582
43		         0.0480
54		         0.0401
63		         0.0353
75		         0.0293
85		         0.0251
95		         0.0214

revircs · Answer

To start, we need to find the order of the reaction. Do you know how to find what order the reaction is?

revircs · Answer

I see that you posted this an hour ago, so ill just go over it for anyone that stumbles upon this.

On your table, you're going to want to make 2 more columns, one labeled ln[A] and the other 1/[A]. Calculate these values by replacing [A] for each concentration value at each time.

From here, we will be able to see which column will have constant differences between the values.

If change in [A] is constant, its 0th order
If change in ln[A] is constant, its 1st order
If change in 1/[A] is constant, its 2nd order

After calculation your two other columns, you will see that change in ln[A] is constant. Therefore, a 1st order reaction.

revircs · Answer

Next, we will need to find the rate constant k. For a first order reaction, the rate = k[A] and the integrated rate law is $$[A] = [A _{o}]e ^{-kt}$$

revircs · Answer

[A] is the concentration at any given time
[Ao] is the initial concentration at t=0
e is a constant
k is the rate constant
t is time

revircs · Answer

To find a concentration at a given time, we first need to solve for the rate constant.

we can do this by plugging in the values we already have from the graph:

  $$0.0825 = 0.0927e ^{-k(8\min)}$$
and solving for k

I calculated it and got k=.015

Since we now have k, we can calculate the concentration after 2 days.

$$[A] = 0.0927e ^{-.015t}$$

Convert 2 days to minutes, and insert that in for t, and you will have your concentration after two days.