Question

The heat of combustion of glucose is given by $C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O$;$\Delta H = - 2840kJ$.Which of the following energy is required for the production of .18 gm of glucose by the reverse reaction.

• A. 28.40 kJ
• B. 2.84 kJ
• C. 5.68 kJ
• D. 56.8 kJ

Answer 1

Answer: (B)

2.84 kJ

Answer 2

$\Delta H_{comb} = - 2840kJ$ for $180g$ of $C_{6}H_{12}O_{6}$

$\therefore\Delta H_{comb}$ for $0.18gm$ glucose

$= \frac{- 2840}{180} \times 0.18 = - 2.84kJ$

For the reverse reaction $\Delta H_{comb}$ of glucose = 2.84 kJ