# Calculating initial concentration from known rate and order of reaction Watch

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For a reaction with the rate equation:

rate =

The reaction is

The reaction is

If the

and

The

What is the

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

rate =

*k*[A][B]^2The reaction is

**first order**with respect to AThe reaction is

**second order**with respect to BIf 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 orderand

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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#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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(Original post by

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?

**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?

**(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

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#5

(Original post by

For a reaction with the rate equation:

rate =

The reaction is

The reaction is

If the

and

The

What is the

**robawalsh**)For a reaction with the rate equation:

rate =

*k*[A][B]^2The reaction is

**first order**with respect to AThe reaction is

**second order**with respect to BIf 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 orderand

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

^{-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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(Original post by

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

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

The actual rate in expt 2 is 7.2 x 10

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

**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}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)

...

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#7

(Original post by

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:

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

...

**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 ]^2But for this I get an answer of 1.92 x 10^-6 (which clearly isn't correct)

...

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}

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(Original post by

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

k = 6.67

Then substituting into experiment 2

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

[B ] = root(7.2 x 10

[B ] = root(7.2 x 10

[B ] =

**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 * 0.03k = 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}Damn Maths :-(

But many thanks for this

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