Question

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?

Answer 1

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.