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[SLEEPINGKING]

Can someone provide full working out for the following question?

The heat of combustion of 2-propanol is 33.5kJ/g. calculate the enthalpy change for the combustion reaction.

 

[snowflakeptical]

Can someone provide full working out for the following question?

The heat of combustion of 2-propanol is 33.5kJ/g. calculate the enthalpy change for the combustion reaction.

I'm not sure if we use mcAt here (not needed)
m= 60.095 (molar mass of propanol)

33.5 kj per g so 60.095, would be 33.5 x 60.095= 2013.1825kj
and add a negative sign

 

[DepressedPenguino]

I got approx. 2010 kJ/mol as my answer but since it is an exothermic reaction (combustion), the change in enthalpy should be negative so -2010kJ/mol

 

[snowflakeptical]

I got approx. 2010 kJ/mol as my answer but since it is an exothermic reaction (combustion), the change in enthalpy should be negative so -2010kJ/mol

how was the working in out? Did you have to find the temp. change first?
What the heck was I doing?

 

[DepressedPenguino]

Btw, I pmed you my working out

 

[DepressedPenguino]

Ehhh it was asking for change in ethalpy not change in Temp (Degrees). I see you got 2013kj/mol? That is kind da close to my ans. Perhaps different periodic tables

 

[snowflakeptical]

Ehhh it was asking for change in ethalpy not change in Temp (Degrees). I see you got 2013kj/mol? That is kind da close to my ans. Perhaps different periodic tables

Ohh I see that I went on to an unnecessary part. I didn't round up until the last bit, maybe that's why. Is my working out similar to yours?

 

[DepressedPenguino]

My ans is quite straight forward:
Simply 33.5kJ/g x 60 (molar mass of propanol according to my periodic table) = - 2010 kJ/mol (exothermic reacttion)

 

[snowflakeptical]

My ans is quite straight forward:
Simply 33.5kJ/g x 60 (molar mass of propanol according to my periodic table) = - 2010 kJ/mol (exothermic reacttion)

Oh what... putting in the 0.095 makes a great difference
Thanks for reminding me about the negative sign

 

[DepressedPenguino]

Oh what... putting in the 0.095 makes a great difference
Thanks for reminding me about the negative sign

No problem and yeah, sometimes doing extra steps might result in an answer that is a couple digits away from the ans that you are suppose to get.

 

[Librah]

Oh what... putting in the 0.095 makes a great difference
Thanks for reminding me about the negative sign

Negative sign isn't needed when calculating heat of combustion calculations, since it's already a given that combustion will be exothermic.

Can someone provide full working out for the following question?

The heat of combustion of 2-propanol is 33.5kJ/g. calculate the enthalpy change for the combustion reaction.

Not entirely sure of this, but i thought heat of combustion was already the enthalpy change? And you've already given it in KJ/g. Unless there's some formal definition that enthalpy changes needs to be KJ/mol.

 

[dan964]

enthalpy change is delta H
so negative.

apart from that exactly as ^

your question is probably wrong, usually molar heat of combustion is per mole. delta h is per gram.

 

[SLEEPINGKING]

It was a question from a past exam paper and the question gives info about heat of combustion not molar

 

[InteGrand]

So is the only thing this question is asking to do to convert kJ/g to kJ/mol ???

 

[Silly Sausage]

Can someone provide full working out for the following question?

The heat of combustion of 2-propanol is 33.5kJ/g. calculate the enthalpy change for the combustion reaction.

Is that the whole question?
For change in enthalpy , you'll need some more information.

 

[snowflakeptical]

converting per gram to per mol requires one step so this question is strange to be asked

 

[Librah]

It was a question from a past exam paper and the question gives info about heat of combustion not molar

Post the exact question, you may have worded it incorrectly.

enthalpy change is delta H
so negative.

apart from that exactly as ^

your question is probably wrong, usually molar heat of combustion is per mole. delta h is per gram.

HSC questions won't actually ask for enthalpy change... they'll word it like "heat of combustion," so you don't need to place the negative sign, since it's assumed combustion will release energy. Though actually for the question being asked by the OP just put a - on the 33.5 KJ/g and you got your answer i suppose.

 

[Librah]

Is that the whole question?
For change in enthalpy , you'll need some more information.

Assuming this is in constant pressure, yer it is equal to that aswell by the way the question is worded there are multiple answers you can come to. Though that value doesn't really mean much since all it gives is the energy lost by combustion/absorbed by the water and doesn't give a "useful" answer.