If $1500 g$ $"1500 g"$ of water was heated by the combustion of $1 g$ $"1 g"$ of sucrose to go from ${25.0}^{\circ} C$ $25.0^@ "C"$ to ${28.3}^{\circ} C$ $28.3^@ "C"$ , what is the enthalpy of combustion in $kJ/mol$ $"kJ/mol"$ ?

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Answer 1 · 2017-05-04T09:12:02

-7080 kJ /mol of sucrose

**Heat released **

$Increase of water temperature = 3.3 K = (28.3-25.0) K$ $"Increase of water temperature = 3.3 K = (28.3-25.0) K"$

$1500 g water = \frac{1500 g}{18.02 g/mol} = 83.24 mol$ $"1500 g water" = "1500 g"/"18.02 g/mol" = "83.24 mol"$ of water

$q = 83.24 mol \cdot {75.4 J\cdotmol}^{-1} K^{-1} \cdot 3.3 K = 20700 J$ $q="83.24 mol" * "75.4 J·mol"^"-1""K"^"-1" *3.3 "K" = "20700 J"$

The amount of heat released by burning 1 g of sucrose is 20.7 kJ per gram of sucrose.

However, 1 mol of sucrose is 342 g.

Therefore $Delta H$ can be computed:

$DeltaH = "-20.7 kJ/g" * "342 g/mol" = "-7080 kJ/mol"$

If "1500 g"1500 g"1500 g" of water was heated by the combustion of "1 g"1 g"1 g" of sucrose to go from 25.0^@ "C"25.0∘C25.0^@ "C" to 28.3^@ "C"28.3∘C28.3^@ "C", what is the enthalpy of combustion in "kJ/mol"kJ/mol"kJ/mol"?