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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