# calculate the initial rate Help

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#1
(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
0
2 years ago
#2
(Original post by Hi freinds)
(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
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.
1
2 years ago
#3
Rate equation or orders of each reactant or units for K so we know the overall order.
0
#4
(Original post by Pigster)
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.
i know great copy and paste you are welcome
0
#5
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 _____________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
Units _________________________________________________________
______________________________________________________________
(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.
______________________________________________________________
______________________________________________________________
(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.
______________________________________________________________
______________________________________________________________
0
2 years ago
#6
(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?
0
#7
(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
0
2 years ago
#8
(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.
0
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