# 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 $k J m o {l}^{- 1}$ which is the data book value for the standard enthalpy of combustion of ethanol?

May 20, 2018

You really should have quoted the specific heat of water in $J \cdot {g}^{-} 1$...

#### Explanation:

We interrogate the combustion reaction...

${C}_{2} {H}_{5} O H \left(l\right) + 3 {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 3 {H}_{2} O \left(l\right) + \Delta$

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

$\text{Moles of ethanol} = \frac{1.21 \cdot g}{46.07 \cdot g \cdot m o {l}^{-} 1} = 0.0263 \cdot m o l$

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

And $\Delta = {m}_{\text{mass of water"xxDeltaTxxunderbrace(4.181*J*K^-1*g^-1)_"specific heat of water}}$

$= 400 \cdot g \times 12.9 \cdot K \times 4.181 \cdot J \cdot {K}^{-} 1 \cdot {g}^{-} 1 = 21574.0 \cdot J$

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

$\Delta {H}_{\text{combustion enthalpy of ethanol}}^{\circ} = \frac{21547.0 \cdot J}{0.0263 \cdot m o l}$

$= - 820.3 \cdot k J \cdot m o {l}^{-} 1$...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.