# Enthalpy change of combustion question

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#1
1.51 g of butan-1-ol heated 300 cm3 of water by 42.0°C. Find the enthalpy change of combustion.
I did q=mc(Delta)T and worked out the moles and everything but I got 19750.5 kJ mol-1 and the answer is supposed to be 2581.1 kJ mol–1.
0
3 years ago
#2
Did you do Delta H = -q / n?
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#3
Not -q/mol haha
I'll show my working

q=mc(delta)T
=300*4.18*(273+42)
=300*4.18*315
=395010/1000 = 395.01 kJ mol-1
Mr Butan-1-ol = 74
mol = 1.51/74 = 0.020405 mol
(delta)H = q/mol = 395.01/0.02 = 19750.5

Also I forgot to put the -ve sign since it's exothermic woops
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3 years ago
#4
(Original post by mydearestpotato)
Not -q/mol haha
I'll show my working

q=mc(delta)T
=300*4.18*(273+42)
=300*4.18*315
=395010/1000 = 395.01 kJ mol-1
Mr Butan-1-ol = 74
mol = 1.51/74 = 0.020405 mol
(delta)H = q/mol = 395.01/0.02 = 19750.5

Also I forgot to put the -ve sign since it's exothermic woops
You don't need to convert to Kelvin.
0
3 years ago
#5
(Original post by mydearestpotato)
1.51 g of butan-1-ol heated 300 cm3 of water by 42.0°C. Find the enthalpy change of combustion.
I did q=mc(Delta)T and worked out the moles and everything but I got 19750.5 kJ mol-1 and the answer is supposed to be 2581.1 kJ mol–1.
E=-mc*delta T
mass= 0.300 kg since volume of water is 300cm^3
c=4.18
change in temperature is 42

E= -0.3*4.18*42= - 52.668 KJ
1.51 g of butanol released 52.668 KJ
1.51 g is how many moles?
moles= 1.51/74=0.0204 moles
1 mole of butanol released 52.668/0.0204= 2581.08 KJ
so enthalpy change is -2581.1 KJ/mol
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#6
Oh I see - thank you!
So canI assume that Kelvin is never used for the q=mc(delta)T questions then?
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