# Rates question

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#1
In this question, the experiment is repeated with double volume- how would this affect the inital rate of reaction?
I have the concs of both reactants, value of K and the old initial rate of reaction

k=5.05 and other values listed in question. Thank you (it's the second part of the question I'm stuck on)
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3 years ago
#2
(Original post by jazz_xox_)
In this question, the experiment is repeated with double volume- how would this affect the inital rate of reaction?
I have the concs of both reactants, value of K and the old initial rate of reaction

k=5.05 and other values listed in question. Thank you (it's the second part of the question I'm stuck on)
The rate of reaction (initial or not) depends upon the concentration of reactants C and D, hence their concentrations are shown in the rate equation.
Repeat the experiment with different starting concentrations and the rate of reaction will be different too - it must be if you look at the rate equation for this reaction.

If you double the volume using just solvent, you will halve the concentrations of C and D. Work those concentrations out and put those values in the rate equation. K will be unchanged as the temperature will be the same in the repeated experiment.
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#3
(Original post by TutorsChemistry)
The rate of reaction (initial or not) depends upon the concentration of reactants C and D, hence their concentrations are shown in the rate equation.
Repeat the experiment with different starting concentrations and the rate of reaction will be different too - it must be if you look at the rate equation for this reaction.

If you double the volume using just solvent, you will halve the concentrations of C and D. Work those concentrations out and put those values in the rate equation. K will be unchanged as the temperature will be the same in the repeated experiment.
Ahh thank you, I have the answer now- I was doubling the concentrations rather than halving them. Could you explain a bit more as to why doubling the volume halves the concentration.. is it to do with concentration = moles / volume.. so if the volume is 2 times as big it will cause the concentration to half?
1
3 years ago
#4
(Original post by jazz_xox_)
Ahh thank you, I have the answer now- I was doubling the concentrations rather than halving them. Could you explain a bit more as to why doubling the volume halves the concentration.. is it to do with concentration = moles / volume.. so if the volume is 2 times as big it will cause the concentration to half?
Yes, that's exactly the reason why
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#5
(Original post by TutorsChemistry)
Yes, that's exactly the reason why
Thank you so much! Do you have a website or any other resources by any chance
0
3 years ago
#6
(Original post by jazz_xox_)
Thank you so much! Do you have a website or any other resources by any chance
Well, yes I have very recently started an A Level chemistry website. The plan is to cover each topic in the A level chemistry syllabuses.

It is very early days and I haven't published about rates yet. So far there are lessons on acids & bases, the pH scale, how to use the ionic product of water. Shortly I will publish about Ka for acids, then Ka for acidic buffers.

Many more topics will follow; eqilibrium and reaction rates are coming soon.
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#7
(Original post by TutorsChemistry)
Well, yes I have very recently started an A Level chemistry website. The plan is to cover each topic in the A level chemistry syllabuses.

It is very early days and I haven't published about rates yet. So far there are lessons on acids & bases, the pH scale, how to use the ionic product of water. Shortly I will publish about Ka for acids, then Ka for acidic buffers.

Many more topics will follow; eqilibrium and reaction rates are coming soon.
Brilliant I'll have a look, thank you
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