Combustion of glucose takes place as
`C_6H_(12)O_6+6O_(2(g)) rarr 6CO_(2(g))+6H_2O_((g))`
`Delta H =- 72` kcal / mole
The energy needed for the production of 1.8 g of glucose by photosynthesis will be 0.82 k cal

Options

A. 0.82 k cal
B. 0.72 k cal
C. 8.2 k cal
D. 7.2 kcal

Answer

B

Explanation

The equation for photosynthesis is the reverse combustion of glucose.
`therefore` `6CO_(2(g)) + 6H_(2)O(g)rarrC_(6)H_(12)O_(6(s))+6O_(2(g)),`
`DeltaH=72` K cal `mol^(-1)`
Molecular mass of glucose = 180 g
180 = 72 K cal
`therefore` for 1.8 g, DeltaH =`(72xx1.8)/(180) = 0.72` k cal