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# Energy Transfer Question watch

1. Hi there!
This is my first post, so plz forgive any mistake that I might make!

anyways I got kinda stuck on this question which says:

In an experiment performed to measure the enthalpy change for the reaction

Cu^2+ + Zn -----> Cu + Zn^2+

3.0 g of zinc powder (an excess) was added to 30.0 cm3 of copper(II) sulfate solutionof concentration 1.00 mol dm–3. The temperature rise of the mixture was 47.6 K.Assuming that the heat capacity of the solution is 4.2 J K–1 g–1, the enthalpy change forthe reaction is given by

A.) Delta
H = -(30 x 4.2 x 47.6)/ 0.03

B.) DeltaH = -(33 x 4.2 x 47.6)/0.03

C.) DeltaH = -(30 x 4.2 x 47.6) x 0.03

D.) DeltaH = -(33 x 4.2 x 47.6) x 0.03
I don't understand either of these answers , I would greatly appreciate any help!
2. (Original post by KL-RC)
Hi there!
This is my first post, so plz forgive any mistake that I might make!

anyways I got kinda stuck on this question which says:

In an experiment performed to measure the enthalpy change for the reaction

Cu^2+ + Zn -----> Cu + Zn^2+

3.0 g of zinc powder (an excess) was added to 30.0 cm3 of copper(II) sulfate solutionof concentration 1.00 mol dm–3. The temperature rise of the mixture was 47.6 K.Assuming that the heat capacity of the solution is 4.2 J K–1 g–1, the enthalpy change forthe reaction is given by

A.) Delta
H = -(30 x 4.2 x 47.6)/ 0.03

B.) DeltaH = -(33 x 4.2 x 47.6)/0.03

C.) DeltaH = -(30 x 4.2 x 47.6) x 0.03

D.) DeltaH = -(33 x 4.2 x 47.6) x 0.03
I don't understand either of these answers , I would greatly appreciate any help!
Hey there, thanks for the follow

Ok so the basis of this calculation is using the equation Q=McΔT, where Q is energy and in this case enthalpy change, M is mass, c is specific heat capacity and ΔT is temp change. So we already have some of these variables
c=4.2 from the question
ΔT=47.6 from the question

All we need to do is calculate the mass, we already have the mass of zinc which is 3g given in the question.

Now you need to work out the mass of copper sulfate solution using Moles=Volume/concentration them moles=mass/MR. The answers are negative because the reaction is exothermic so energy is given off

If you need any more help just message me
3. Thank you so very much for the reply!!! I'm very greatful!!

just one thing though I understand why the symbol is negative (because the reactions exothermic) but for the mass of copper sulphate wasn't the formula Mole = Concentration x Volume?? :O
4. (Original post by Schrödingers Cat)
Now you need to work out the mass of copper sulfate solution using Moles=Volume/concentration them moles=mass/MR.
This is wrong.
We assume the solution has the same density as water which is 1g/cm3
So volume in cm3 = mass in g
5. (Original post by KL-RC)
Thank you so very much for the reply!!! I'm very greatful!!

just one thing though I understand why the symbol is negative (because the reactions exothermic) but for the mass of copper sulphate wasn't the formula Mole = Concentration x Volume?? :O
Yes you're right I was thinking conc=mol/vol is how I always remembered it

Oh, the bit I missed off . It's about the zinc being in excess so not all of it reacts so work out the moles of copper sulphate reacting and factor it in accordingly

This is wrong.
We assume the solution has the same density as water which is 1g/cm3
So volume in cm3 = mass in g

For mass of sulphate ^^^^

This is wrong.
We assume the solution has the same density as water which is 1g/cm3
So volume in cm3 = mass in g
Yep my bad, haven't done chemistry in 5 months
7. (Original post by Schrödingers Cat)
Yep my bad, haven't done chemistry in 5 months

Oh...haha its alright Thank you very much though!!!
8. (Original post by KL-RC)
Oh...haha its alright Thank you very much though!!!
Are you sorted now?
9. (Original post by KL-RC)
Oh...haha its alright Thank you very much though!!!
You're welcome, my last edit should help you complete the rest of the question
Are you sorted now?
Yea almost I just PM you my last question!! sorry!!!
11. Why would Q be the enthalpy change? Surely you should have dQ=mcdT = 33g x 4.2 x 47.6. Then you would have to divide by 1000 and further divide by the moles of solution? Or am I misunderstanding something.
12. (Original post by Protoxylic)
Why would Q be the enthalpy change? Surely you should have dQ=mcdT = 33g x 4.2 x 47.6. Then you would have to divide by 1000 and further divide by the moles of solution? Or am I misunderstanding something.
Q is the energy transferred in Joules and so is the enthalpy change for the given amount of reactants. Divide by moles of which ever reactant is not in excess gives the enthalpy change in Jmol-1.
What you are suggesting gives the enthalpy change in kJmol- which would be what you would expect but it isn't one of the choices.
Q is the energy transferred in Joules and so is the enthalpy change for the given amount of reactants. Divide by moles of which ever reactant is not in excess gives the enthalpy change in Jmol-1.
What you are suggesting gives the enthalpy change in kJmol- which would be what you would expect but it isn't one of the choices.
I understand 2 of those answers are quoted in Jmol^-1. But why would Q be equal to delta H? I think I'm getting confused between enthalpy change of reaction and just enthalpy change. Is this just a consequence of the first law of thermodynamics? Since the internal energy change would just be equal to the thermal energy change which is equal to the enthalpy change at constant pressure.
deltaH is per mol (either Jmol-1 or kJmol-1) so by dividing by the number of moles that reacted Q becomes delta H.
That is what I am saying. On one of the other posts he/she said that Q=delta H. I understand that it would be deltaH if it was just an enthalpy change, but an enthalpy change of reaction is per mol.
15. (Original post by Protoxylic)
That is what I am saying. On one of the other posts he/she said that Q=delta H. I understand that it would be deltaH if it was just an enthalpy change, but an enthalpy change of reaction is per mol.
Sorry I decided to delete my post as I realised you were asking something else but you posted whilst I was deleting!

I agree with what you say here. The Q= enthalpy change comment was made by another poster not the OP. Enthalpy change is per mole hence the dividing by 0.03 in the question.
16. Hey there if anyone recently is wanting to know how to do the question here is how!

first of all the equation for determining enthalpy change of reaction. which is always given in Kj mol^-1
which is kj divided by mols:
Kj/mols
when you bring the mols above in maths the power is multiplied by -1
therefore it is Kj mol^-1
so first of all we find the energy transferred which is the equation Q=McΔT the we divide it by the moles of what we listed for the mass in our Q=McΔT equation

1)
mass x specific heat capacity x change in temperature = KJ (energy transferred)
(mass is always in cm^3 if we are talking about solutions as 1g=1cm)
(30 x 4.12 x 47.6)

2) find the moles of what you listed for mass. in this case we listed copper sulphate solution
to find out moles it is C x V (note if volume is given in cm^3 you should divide by 1000 to convert to dm^3)
moldm^-3 x (cm/1000) =
1.00 moldm^-3 x (30/1000) = 1.00 x 0.03 = 0.03 = mols

therefore the overall equation to give the Kj mol^-1 (Q=McΔT)/mols =

would be
-(30 x 4.12 x 47.6) divided by 0.03
minus sign since this is an exothermic reaction!!

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