Calculating initial concentration from known rate and order of reactionWatch

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Thread starter 6 years ago
#1
For a reaction with the rate equation:

rate = k[A][B]^2

The reaction is first order with respect to A
The reaction is second order with respect to B

If the initial concentration [A] in experiment 1 is 0.060 mol dm^-3 and 0.040 mol dm^-3 in experiment 2, and A is first order

and

The initial rate in experiment 1 is 3.6 x 10^-4 mol dm^-3 s^-1 and 7.2 x 10^-4 mol dm^-3 s^-1 in experiment 2...

What is the initial concentration of B in experiment 2, given that the initial rate in experiment 1 is 0.030 mol dm^-3?

I understand how to calculate these for the most part, but I'm not getting the maths behind this one. Could somebody please give me the concentration and a brief explanation of the maths behind it?

Thank you
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Thread starter 6 years ago
#2
70 views and no reply 0
6 years ago
#3
I'm really tired so this answer might make no sense but is it maybe that you have to work out k in the first reaction? Cause you have [A], the rate and [B]. Then k will stay the same at a constant temperature so you can use k, the rate and [A] to work out [B] in the second reaction?
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Thread starter 6 years ago
#4
(Original post by NinjaNerdfighter)
I'm really tired so this answer might make no sense but is it maybe that you have to work out k in the first reaction? Cause you have [A], the rate and [B]. Then k will stay the same at a constant temperature so you can use k, the rate and [A] to work out [B] in the second reaction?
No, it should be something like [A] is first order so if [A] is x 2/3 then the rate is x 4/3 (?) but [B] must increase/decrease by (a certain) amount

(we know [B] is second order so doubling it would result in the rate increasing by a factor of 8, for example)

...or something 0
6 years ago
#5
(Original post by robawalsh)
For a reaction with the rate equation:

rate = k[A][B]^2

The reaction is first order with respect to A
The reaction is second order with respect to B

If the initial concentration [A] in experiment 1 is 0.060 mol dm^-3 and 0.040 mol dm^-3 in experiment 2, and A is first order

and

The initial rate in experiment 1 is 3.6 x 10^-4 mol dm^-3 s^-1 and 7.2 x 10^-4 mol dm^-3 s^-1 in experiment 2...

What is the initial concentration of B in experiment 2, given that the initial rate in experiment 1 is 0.030 mol dm^-3?
I understand this to be a typo and should say concentration

I understand how to calculate these for the most part, but I'm not getting the maths behind this one. Could somebody please give me the concentration and a brief explanation of the maths behind it?

Thank you
Given that [A] is first order you would expect the rate to reflect the change in [A]. It decreases from 0.06 to 0.04 so the rate should do the same, i.e. decrease from 3.6 x 10-4 to 2.4 x 10-4

Therefore any difference in the rate is due to [B ]

The actual rate in expt 2 is 7.2 x 10-4, meaning a threefold increase.

As B is second order this means that [B ] must have increased by a factor of root(3) = 1.73.

It was 0.03, hence it is now 0.03 x 1.73 = 0.052 mol dm-3
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Thread starter 6 years ago
#6
(Original post by charco)
I understand this to be a typo and should say concentration

Given that [A] is first order you would expect the rate to reflect the change in [A]. It decreases from 0.06 to 0.04 so the rate should do the same, i.e. decrease from 3.6 x 10-4 to 2.4 x 10-4

Therefore any difference in the rate is due to [B ]

The actual rate in expt 2 is 7.2 x 10-4, meaning a threefold increase.

As B is second order this means that [B ] must have increased by a factor of root(3) = 1.73.

It was 0.03, hence it is now 0.03 x 1.73 = 0.052 mol dm-3
These values seem to work.

However when I try to calculate [B] using the equation after calculating the rate constant, it doesn't fit

k = ([0.02] x [0.02]^2) / 1.2 x 10^-4 = 0.0666666 mol^-2 dm^6
(Values from another relevant experiment. This value for k fits with other experimental data so must be correct)

So surely I would expect that:

rate (of expt. 2) x 0.066 mol^-2 dm^6 (k) x 0.04 ([A]) = [B ]^2

But for this I get an answer of 1.92 x 10^-6 (which clearly isn't correct)

...
0
6 years ago
#7
(Original post by robawalsh)
These values seem to work.

However when I try to calculate [B] using the equation after calculating the rate constant, it doesn't fit

k = ([0.02] x [0.02]^2) / 1.2 x 10^-4 = 0.0666666 mol^-2 dm^6
(Values from another relevant experiment. This value for k fits with other experimental data so must be correct)

So surely I would expect that:

rate (of expt. 2) x 0.066 mol^-2 dm^6 (k) x 0.04 ([A]) = [B ]^2

But for this I get an answer of 1.92 x 10^-6 (which clearly isn't correct)

...
I've just done the maths and it works out ....

Using the data from experiment 1 to find the rate constant k

k = rate/[A][B ][B ]

k = 3.6 x 10-4 / 0.06 * 0.03 * 0.03

k = 6.67

Then substituting into experiment 2

[B ] = root{rate /(k * [A])}

[B ] = root(7.2 x 10-4/6.67 * 0.04)

[B ] = root(7.2 x 10-4/6.67 * 0.04)

[B ] = 0.052 mol dm-3
0
Thread starter 6 years ago
#8
(Original post by charco)
I've just done the maths and it works out ....

Using the data from experiment 1 to find the rate constant k

k = rate/[A][B ][B ]

k = 3.6 x 10-4 / 0.06 * 0.03 * 0.03

k = 6.67

Then substituting into experiment 2

[B ] = root{rate /(k * [A])}

[B ] = root(7.2 x 10-4/6.67 * 0.04)

[B ] = root(7.2 x 10-4/6.67 * 0.04)

[B ] = 0.052 mol dm-3
Ahh, I got the units wrong and forgot to do the roots.

Damn Maths :-(

But many thanks for this
0
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