Question

# The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 25oC are −400 kJ/mol, −300 kJ/mol and −1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25oC is:

A
+2900 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2900 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
16.11 kJ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
+16.11 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C −16.11 kJSolution:- (C) −16.11kJCombustion of glucose-C6H12O6(s)+6O2(g)⟶6CO2(g)+6H2O(l)ΔH0=∑ΔH0f(product)−∑ΔH0f(reactant)∴ΔH0=(6×(−400)+6×(−300))−(−1300+0)⇒ΔH0=−4200+1300=−2900kJ/molNow,ΔH0(kJ/gm)=ΔH0(kJ/mol)Mol. wt.(in gm)Molecular weight of glucose =180gm∴ΔH0(kJ/gm)=−2900180=−16.11kJ/gmHence the standard enthalpy of combustion per gram of glucose at 25℃ is −16.11kJ.

Suggest Corrections
0
Join BYJU'S Learning Program
Select...
Related Videos
Thermochemistry
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program
Select...