Question

The heat evolved in the combustion of glucose is shown in the equation :

C₆H₁₂O₆ + 6O₂ (g) → 6CO₂ (g) + 6 H₂O (g), Δ_c H

= – 2840 kJ mol^–1

What is the energy requirement for production of 0.36 g of glucose by the reverse reaction ?

Answer 1

The given equation is :

C₆H₁₂O₆ + 6O₂ (g) → 6CO₂ (g) + 6H₂O (g) ;

Δ_rH = – 2840 kJ mol^–1 

Writing the reverse reaction, we have

6CO₂ (g) + 6H₂O(g) → C₆H₁₂O₆ (s) + 6O₂ (g) ;

Δ_r H = + 2840 kJ mol^–1 

Thus, for production of 1 mole of C₆H₁₂O₆ (= 72 + 12 + 96 = 180 g), heat required (absorbed) = 2840 kJ.

∴ For production of 0.36 g of glucose, heat absorbed 

= (2840/180 x 0.36) = 5.68 KJ