Enthalpy chnages
If an experiment is carried out to calculate the enthalpy change of combustion of hexane using 200cm3 of water, how would the value differ if the experiment was repeated but instead with 250cm3 of water?
Also would appreciate as to why it changes/ doesn't.
Also would appreciate as to why it changes/ doesn't.
In theory, it wouldn't.
In practice I'd guess you'd get a more exothermic value as the water would heat up less (assuming the same heat transfer, i.e. heating time) and therefore there'd be less heat loss.
Then again, if you chose to heat up both water volumes by the same temperature change would lead to a less exothermic value with the larger volume of water because it would take longer to heat up and there would be more time for heat to he lost.
w00t
In practice I'd guess you'd get a more exothermic value as the water would heat up less (assuming the same heat transfer, i.e. heating time) and therefore there'd be less heat loss.
Then again, if you chose to heat up both water volumes by the same temperature change would lead to a less exothermic value with the larger volume of water because it would take longer to heat up and there would be more time for heat to he lost.
w00t
There are couple of ways to determine the ΔH of a reaction. Q = MCΔT would be the suitable method to determine the ΔH for this example. Remember that Q is the measure of energy transferred in Joules. And this can be converted to ΔH by converting Q (J) into kJ/mol by taking into account the number of mol of hexane that was responsible for the energy transferred.
As Q takes into account M, the mass of water, an increase in the volume or mass of water will result in an increase in Q and thus resulting in a bigger ΔH value. In this example, a greater exothermic value will result.
For other scenarios, the different methods to directly determine the ΔH would include:
- Hess's Cycle
- Born-Haber Cycle
- Bond Enthalpy
- Enthalpy Profile Diagram, with calibrated y-axis
I wish you all the best with your studies.
As Q takes into account M, the mass of water, an increase in the volume or mass of water will result in an increase in Q and thus resulting in a bigger ΔH value. In this example, a greater exothermic value will result.
For other scenarios, the different methods to directly determine the ΔH would include:
- Hess's Cycle
- Born-Haber Cycle
- Bond Enthalpy
- Enthalpy Profile Diagram, with calibrated y-axis
I wish you all the best with your studies.
Original post by Pigster
In theory, it wouldn't.
In practice I'd guess you'd get a more exothermic value as the water would heat up less (assuming the same heat transfer, i.e. heating time) and therefore there'd be less heat loss.
Then again, if you chose to heat up both water volumes by the same temperature change would lead to a less exothermic value with the larger volume of water because it would take longer to heat up and there would be more time for heat to he lost.
w00t
In practice I'd guess you'd get a more exothermic value as the water would heat up less (assuming the same heat transfer, i.e. heating time) and therefore there'd be less heat loss.
Then again, if you chose to heat up both water volumes by the same temperature change would lead to a less exothermic value with the larger volume of water because it would take longer to heat up and there would be more time for heat to he lost.
w00t
but why? i don't understand if you use Q = MCΔT then the only thing that increases is the mass of water so surely the
nvm i understand now
Original post by dongseuk
There are couple of ways to determine the ΔH of a reaction. Q = MCΔT would be the suitable method to determine the ΔH for this example. Remember that Q is the measure of energy transferred in Joules. And this can be converted to ΔH by converting Q (J) into kJ/mol by taking into account the number of mol of hexane that was responsible for the energy transferred.
As Q takes into account M, the mass of water, an increase in the volume or mass of water will result in an increase in Q and thus resulting in a bigger ΔH value. In this example, a greater exothermic value will result.
As Q takes into account M, the mass of water, an increase in the volume or mass of water will result in an increase in Q and thus resulting in a bigger ΔH value. In this example, a greater exothermic value will result.
If the same amount of fuel is burnt, the amount of energy released, Q, would be the same. A larger mass of water, M, would result in a smaller temperature change, ΔT, so ΔH would be the same.If you forced ΔT to be the same with a larger mass of water, Q would, as you said be increased. But to make ΔT the same, you'd have to burn more hexane and hence n would be larger, so ΔH = Q / n. Both Q and n are increased by the same degree, so ΔH is the same.
Original post by thefatone
but why? i don't understand if you use Q = MCΔT then the only thing that increases is the mass of water so surely the
nvm i understand now
nvm i understand now
It's good that now you understand.
Just to clear your doubts, the increase in the mass of water would result in an increase in Q so ultimately the ΔH value.
Good luck with your studies.
Original post by dongseuk
It's good that now you understand.
Just to clear your doubts, the increase in the mass of water would result in an increase in Q so ultimately the ΔH value.
Good luck with your studies.
Just to clear your doubts, the increase in the mass of water would result in an increase in Q so ultimately the ΔH value.
Good luck with your studies.
no not that xD
the bit where it enthalpy change shouldn't change in theory is the bit i didn't understand but now i do
Original post by dongseuk
Just to clear your doubts, the increase in the mass of water would result in an increase in Q so ultimately the ΔH value.
ΔcH for liquid hexane = -4163 kJ mol-1.
If ΔcH increased with an increase the mass of water being heated, how could there be only one value be in a databook?
ΔH is, to be clear, unaffected by the mass of water being heated.
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