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Standard XII

Chemistry

Enthalpy of Combustion

When 0.36 g o...

Question

When 0.36 g of glucose was burned in a bomb calorimeter (Heat capacity $600 J K^{- 1}$ ) the temperature rise by 10 K. Calculate the standard molar enthalpy of combustion (MJ/mole).

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Solution

$M o l e s o f s u b s t a n c e = \frac{g i v e n m a s s}{M o l a r m a s s}$

$M o l e s o f g l u c o s e = \frac{0.36}{180} = 0.002 m o l e s$
$M o l a r e n t h a l p y o f c o m b u s t i o n = \frac{h e a t c a p a c i t y \times r i s e i n t e m p e r a t u r e}{m o l e s o f g l u c o s e}$
$M o l a r e n t h a l p y o f c o m b u s t i o n = \frac{600 \times 10}{0.002} = 3 \times 10^{6} J / m o l = 3 M J / m o l$

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When 0.36 g of glucose was burned in a bomb calorimeter (Heat capacity 600JK−1) the temperature rise by 10 K. Calculate the standard molar enthalpy of combustion (MJ/mole).

Moles of substance=given massMolar massMoles of glucose=0.36180=0.002 molesMolar enthalpy of combustion=heat capacity×rise in temperaturemoles of glucoseMolar enthalpy of combustion=600×100.002=3×106 J/mol=3 MJ/mol

When 0.36 g of glucose was burned in a bomb calorimeter (Heat capacity $600 J K^{- 1}$ ) the temperature rise by 10 K. Calculate the standard molar enthalpy of combustion (MJ/mole).