calculate the initial rate Help

(iv) In a second experiment at the same temperature, water was added to the originalreaction mixture so that the total volume was doubled.
Calculate the initial rate of reaction in this second experiment.
data from previous question
(iii) When the initial concentration of X was 0.024 mol dm–3 and the initial concentration of hydroxide ions was 0.035 mol dm–3, the initial rate of the reaction was
8.5 × 10–5 mol dm–3 s–1.
rate constant = 0.1

Reply 1

Pigster

Original post by Hi freinds

Nice copy and pasting skillz.

With the information provided, I believe it is impossible for us to help you. Had you provided us with a reaction and the unit for the rate constant... possibly. With a rate equation, certainly. But no.

Reply 2

username1445490

More info needed.
Rate equation or orders of each reactant or units for K so we know the overall order.

Reply 3

Hi freinds

Original post by Pigster

i know great copy and paste you are welcome

Reply 4

Hi freinds

Original post by Madasahatter

More info needed.
Rate equation or orders of each reactant or units for K so we know the overall order.

here is the full question
The rate of hydrolysis of an ester X (HCOOCH2CH2CH3) was studied in alkaline conditions at a given temperature. The rate was found to be first order with respect to the ester and first order with respect to hydroxide ions.
(iii) When the initial concentration of X was 0.024 mol dm–3 and the initial concentration of hydroxide ions was 0.035 mol dm–3, the initial rate of the reaction was
8.5 × 10–5 mol dm–3 s–1.
Calculate a value for the rate constant at this temperature and give its units.
Calculation _____________________________________________________
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Units _________________________________________________________
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(3)
(iv) In a second experiment at the same temperature, water was added to the originalreaction mixture so that the total volume was doubled.
Calculate the initial rate of reaction in this second experiment.
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(1)
(v) In a third experiment at the same temperature, the concentration of X was half that used in the experiment in part (a) (iii) and the concentration of hydroxide ions was three times the original value.
Calculate the initial rate of reaction in this third experiment.
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Reply 5

Pigster

Original post by Hi freinds

here is the full question
blah blah blah

It says that it is first order WRT both reactants.
It says you are doubling the volume, so what does it do to both concs? And hence to the rate?

Reply 6

Hi freinds

Original post by Pigster

It says that it is first order WRT both reactants.
It says you are doubling the volume, so what does it do to both concs? And hence to the rate?

lol whats with the bla bla
and if you double rate you halve conc i think

Reply 7

Pigster

Original post by Hi freinds

lol whats with the bla bla
and if you double rate you halve conc i think

Simple GCSE rates: double the conc = double the number of particles in a given volume = more collisions per second = faster rate.
A level rates: double the conc = double the number of particles in a given volume = (if 1st order) double the collisions per second = double the rate.

Reply 8

1234kelly

Original post by Pigster

i get that the volume doubles but they only doubled the conc of the water not the easter as well so it can't just be double the rate ?

Reply 9

GeT_iN_SHinJI

iii) The rate has an order of 1 with respect to both the ester and hydroxide ions so you can assume the equation is: rate = k[X][OH-]. After that it's simply substituting the initial concentrations and rate into the equation. So 8.5 × 10–5 = k(0.024)(0.035) and rearrange to find k. Also don't forget to calculate the units.

iv) If you double the volume of water, the concentrations will halve (since the solutions have become more dilute) and therefore the rate will be divided by 4.

v) The concentration of X is halved, so the rate is halved. The concentration of hydroxide ions is multiplied by 3 so the rate is multiplied by 3. So overall the rate is multiplied by 1.5 (0.5 x 3).

Reply 10

1234kelly

Original post by GeT_iN_SHinJI

For Iv) for divide by 4 and not 2? Sorry if it’s a silly question

Reply 11

1234kelly

Original post by GeT_iN_SHinJI

Also the hydroxide ions are from the water during the hydrolysis so if we’re adding more water why does the concentration of OH decrease ? I know your answer is completely right I just don’t understand why

Reply 12

1234kelly

wait never mind i can't believe i made such a silly mistake , the OH is not from the water. The OH is from the alkaline conditions but it acts as a cataylst for the hydrolysis no? so why is it in the rate reaction as I thought we don't include catalysts in the rate reaction

Reply 13

GeT_iN_SHinJI

Original post by 1234kelly

For Iv) for divide by 4 and not 2? Sorry if it’s a silly question

If the concentration of OH- is halved then the rate is halved. If the concentration of X is halved then the rate is halved. Therefore overall the rate is divided by 4 (rate / 2 / 2). I'm actually not really sure on this one but that was my take.