The heat evolved in the combustion of glucose is shown in the equation : C_(6)H_(12)O_(6)(s) + 6O_(2)(g) rarr 6CO_(2)(g) + 6H_(2)O (g) , Delta_(c ) H = -2840 kJ mol^(-1) What is the energy requirement for production of 0.36g of glucose by the reverse reaction ?

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

1.	The heat evolved in the combustion of glucose is shown in the equation : C_(6)H_(12)O_(6)(s) + 6O_(2)(g) rarr 6CO_(2)(g) + 6H_(2)O (g) , Delta_(c ) H = -2840 kJ mol^(-1) What is the energy requirement for production of 0.36g of glucose by the reverse reaction ?
Answer» Solution :The given equation is `:C_(6)H_(12)O_(6)+ 6O_(2)(G) rarr6CO_(2)(g) +6H_(2)O(g), Delta_(r)H= -2849 k J MOL^(-1)` WRITTEN the reverse reaction, we have `6CO_(2)(g)+6H_(2)O(g) rarr C_(6)H_(12)O_(6)(s)+ 6O_(2)(g), Delta_(r)H= + 2840 kJ mol^(_1)` Thus, for production of 1 mole of `C_(6)H_(12)O_(6)( = 72 + 12 + 96= 180 g )`, heat required( absorbed) `= 2840kJ`ltbgt `:. `For production of0.36 g of glucose , heat absorbed`=(2860)/(180) XX 0.36 = 5.68 kJ`