Enthalpy calculation

OP

Original post by Pigster

I'll bet you've done loads of examples in class. What are you thoughts on how to do it?

Actually the situation is that my teachers has gone to visit other school to see how their pupils are doing on their practicals and the few days that my teacher was teaching , we were doing practicals so he told our class that we can do practicals now since you can learn the theory by yourself which is reasonable but i'm finding this theory super difficult and i don't really understand much.

This is basically a new topic i know almost nothing about but i know basic formulas such as moles=mass/mr
n=v*c and q=mc▲t

we have done one lesson on calculating bond enthalpies knowing what exothermic and endothermic reactions are but that's about it so please walk me through how to do this.

Reply 3

You have to use that equation q=mc▲t. You know C = 4.18, you convert degrees to Kelvin by adding 273 and your mass if the amount of CuSO4. (50cm3)
You get the answer and use the formula H= Q/mol
You work out the mol of CuSO4 and then you have the answer.
You have to divide the answer b y 1000 to get your final answer in KJ. :smile:

(I think that's how you do it but I might be wrong)

Reply 4

OP

Original post by yorobun

You have to use that equation q=mc▲t. You know C = 4.18, you convert degrees to Kelvin by adding 273 and your mass if the amount of CuSO4. (50cm3)
You get the answer and use the formula H= Q/mol
You work out the mol of CuSO4 and then you have the answer.
You have to divide the answer b y 1000 to get your final answer in KJ. :smile:

(I think that's how you do it but I might be wrong)

i'm pretty sure that it doesn't matter whether change in temperature is in kelvin or degrees Celsius

This is a new formula what are the letter standing for?

Reply 5

H is the enthalpy change, Q is the energy and mol is just moles. :smile:

You might need to change cm3 to dm3

(edited 8 years ago)

Reply 6

OP

Original post by yorobun

You have to use that equation q=mc▲t. You know C = 4.18, you convert degrees to Kelvin by adding 273 and your mass if the amount of CuSO4. (50cm3)
You get the answer and use the formula H= Q/mol
You work out the mol of CuSO4 and then you have the answer.
You have to divide the answer b y 1000 to get your final answer in KJ. :smile:

(I think that's how you do it but I might be wrong)

But i already did? that's what i did in the first step to finding the mass of CuSO4

Reply 7

The mass you need for the equation is the mass of the surrounding, the surrounding is the CuSO4 so if you think about it like as if you were doing the experiment, you would pour the zinc powder in the solution of CuSO4 which is 50cm3 ( 0.05dm3). I get confused at this bit aswell but I remember doing a question and I had to use 100cm3 instead of actually working out the mass. Sorry if it doesn't make sense :s-smilie:

Reply 8

Original post by yorobun

You have to use that equation q=mc▲t. You know C = 4.18, you convert degrees to Kelvin by adding 273 and your mass if the amount of CuSO4. (50cm3)
You get the answer and use the formula H= Q/mol
You work out the mol of CuSO4 and then you have the answer.
You have to divide the answer b y 1000 to get your final answer in KJ. :smile:

(I think that's how you do it but I might be wrong)

You don't need to convert temp to kelvin. Let x be initial temp in degrees C and y be end temp in degrees c, so delta T =y-x. In kelvin, delta T =(y+273) -(x+273) =y-x

Reply 9

Original post by samb1234

You don't need to convert temp to kelvin. Let x be initial temp in degrees C and y be end temp in degrees c, so delta T =y-x. In kelvin, delta T =(y+273) -(x+273) =y-x

So its doesn't matter? Thankyou xD

Reply 10

Original post by yorobun

So its doesn't matter? Thankyou xD

No it doesn't matter for anything involving temperature change as the +273 will cancel

Reply 11

Original post by thefatone

Zinc will displace copper from copper (II) sulphate solution according to the following equation:

CuSO₄(aq) + Zn(s) --> Cu(s) + ZnSO₄(aq)
Ifan excess of zinc powder is added to 50 cm³ of 1.0moldm^-3 copper(II) sulphate, the temperature increases by 6.3 ^oC. Calculate the enthalpy change for the reaction.

Assuming ρ =1.0 gcm^-3 and c = 4.18 Jg^-1K^-1

^{someone pls talk me through how to do this}

Have you done this yet or do you still need help?

Reply 12

OP

Original post by yorobun

The mass you need for the equation is the mass of the surrounding, the surrounding is the CuSO4 so if you think about it like as if you were doing the experiment, you would pour the zinc powder in the solution of CuSO4 which is 50cm3 ( 0.05dm3). I get confused at this bit aswell but I remember doing a question and I had to use 100cm3 instead of actually working out the mass. Sorry if it doesn't make sense :s-smilie:

Original post by samb1234

Have you done this yet or do you still need help?

i don't understand

here's what i've done
n=(v*c)/1000
=0.05 mol of CuSO4

Mass=mol*mr
= 0.05*(63.5+32.1+16*4)
=7.98g

q=mc▲t
=7.98*4.18*6.3
=210.15J

h=q/mol
=210.15/0.05
=4202.9

4202.9/1000=4.203kJ/mol
of course -4.203kJ/mol bc this is an exothermic reaction
but this isn't the correct answer

Reply 13

OP

Original post by samb1234

Have you done this yet or do you still need help?

no i've done nothing but get wrong answers so far so any help is appreciated

Reply 14

Original post by thefatone

Zinc will displace copper from copper (II) sulphate solution according to the following equation:

CuSO₄(aq) + Zn(s) --> Cu(s) + ZnSO₄(aq)
Ifan excess of zinc powder is added to 50 cm³ of 1.0moldm^-3 copper(II) sulphate, the temperature increases by 6.3 ^oC. Calculate the enthalpy change for the reaction.

Assuming ρ =1.0 gcm^-3 and c = 4.18 Jg^-1K^-1

^{someone pls talk me through how to do this}

Step 1: we firstly need to work out what the energy change for the reaction is. We therefore can apply the formula Q=mc delta T. We know delta T, we are given c in the question but we now need to calculate the mass. The mass we use is the mass of the overall solution, and we are told in the question that the density of the solution is 1gcm-3, so the overall mass of the solution is therefore 50g. We can then plug the numbers in to get a value for the energy released in J.

Step 2: We know that the unit for enthalpy change is KJ/mol , so we firstly need to go about changing our value from the first part into KJ by dividing by 1,000. We then calculate the number of moles of CuSO4 that reacted, and divide our energy change in KJ by the number of moles to get our final answer

Reply 15

OP

2.1 Exercise 1 - measuring enthalpy changes.doc27.1KB

Original post by samb1234

Step 1: we firstly need to work out what the energy change for the reaction is. We therefore can apply the formula Q=mc delta T. We know delta T, we are given c in the question but we now need to calculate the mass. The mass we use is the mass of the overall solution, and we are told in the question that the density of the solution is 1gcm-3, so the overall mass of the solution is therefore 50g. We can then plug the numbers in to get a value for the energy released in J.

Step 2: We know that the unit for enthalpy change is KJ/mol , so we firstly need to go about changing our value from the first part into KJ by dividing by 1,000. We then calculate the number of moles of CuSO4 that reacted, and divide our energy change in KJ by the number of moles to get our final answer

maybe i should've clarified
In all the following questions, assume that thedensities and specific heat capacities of the solutions are the same as purewater i.e. ρ = 1.0 gcm^-3 and c = 4.18 Jg^-1K^-1

Reply 16

Original post by thefatone

maybe i should've clarified
In all the following questions, assume that thedensities and specific heat capacities of the solutions are the same as purewater i.e. ρ = 1.0 gcm^-3 and c = 4.18 Jg^-1K^-1

Clarify what? I know what you just said, and I used it in the solution

Reply 17

OP