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Question: ΔHc of butan-1-ol?
Here is the question, "An experiment was carried out to determine the enthalpy of combustion of butan-1-ol (CH3CH2CH2CHOH). 1.33g of butan-1-ol was burned. The heat produced caused a temperature increase of 22.9°C in 500cm3 of water. Given that the specific heat capacity of water is 4.2Jg-1°C-1, its density 1g cm-3 and that specific heat capacity of any other apparatus, is negligible, calculate the enthalpy change of combustion of butan-1-ol." Mr of butan-1-ol = 74
Ok, I have tried some workings but they don't correspond with the trend of enthalpies of combustion of Methanol, Ethanol, Propan-1-ol and Pentan-1-ol which I have been given. Could someone explain this for me please? Much obliged.
Ok, I have tried some workings but they don't correspond with the trend of enthalpies of combustion of Methanol, Ethanol, Propan-1-ol and Pentan-1-ol which I have been given. Could someone explain this for me please? Much obliged.
Reply 1
Hey, im not sure if this is right i did chem aggggeeees agoo... but here goes.
Use the equation Q=M x C x change in temp
So M the mass is 0.5, 500 of water converted into kg
C is 4.2
Temp change is 22.9
Q= 0.5 x 4.2 x 22.9
Q= 48.09KJ
Moles of butan.... burned is mass/Mr
= 1.33/74
= 0.0180
Enthalpy change is.... -(Heat energy/Moles of butan...)
=-(48.09/0.0180)
= 2671.7KJmol-1
Hope thats ryt... i know the set out is pooor.. sorry
Use the equation Q=M x C x change in temp
So M the mass is 0.5, 500 of water converted into kg
C is 4.2
Temp change is 22.9
Q= 0.5 x 4.2 x 22.9
Q= 48.09KJ
Moles of butan.... burned is mass/Mr
= 1.33/74
= 0.0180
Enthalpy change is.... -(Heat energy/Moles of butan...)
=-(48.09/0.0180)
= 2671.7KJmol-1
Hope thats ryt... i know the set out is pooor.. sorry
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