1.21 g of ethanol, C_2H_5OH, was burned in a spirit burner. The heat produced raised the temperature of 400 g of water placed in a beaker above the flame from 17.0 °C to 29.9 °C. How do you calculate the enthalpy change for the reaction taking place?

Why is this value not equal to -1371 kJmol^(-1) which is the data book value for the standard enthalpy of combustion of ethanol?

1 Answer
anor277
May 20, 2018

You really should have quoted the specific heat of water in J*g^-1...

Explanation:

We interrogate the combustion reaction...

C_2H_5OH(l) + 3O_2(g) rarr 2CO_2(g)+3H_2O(l)+Delta

So two questions: is mass conserved; is charge conserved? For both questions the answer is clearly yes...and we operate on this basis.

"Moles of ethanol"=(1.21*g)/(46.07*g*mol^-1)=0.0263*mol

Now this site quotes the specific heat of water as 4.181*J*K^-1*g^-1, and you should have dug for this and not I.

And Delta=m_"mass of water"xxDeltaTxxunderbrace(4.181*J*K^-1*g^-1)_"specific heat of water"

=400*gxx12.9*Kxx4.181*J*K^-1*g^-1=21574.0*J

But this heat output was associated with the combustion of the GIVEN MOLAR quantity.... i.e.

DeltaH_"combustion enthalpy of ethanol"^@=(21547.0*J)/(0.0263*mol)

=-820.3*kJ*mol^-1...now this value is LESS than half of the quoted value. Why the shortfall? Well we have assumed that hear transfer was perfect. I would argue that is not. Heat is lost to the surroundings, and heat is lost to the apparatus.

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