The enthalpy change involved in the oxidation of glucose is `-2880kJ mol^(-1)`. Twenty five per cent of this energy is available for muscular work . If `100kJ` of muscular work is needed to walk one kilometre, what is the maximum distance that a person will be able to walk after eating `120g` of glucose ?

Explanation

Molar mass of glucose `(C_(6)H_(12)O_(6))`
`=(6 xx 12 +12 xx1xx6 xx 16) g mol^(-1)`
`= 180 g mol^(-1)`
`C_(6)H_(12)O_(6)(s) rarr 6CO_(2)(g) +6H_(2)O(l) DeltaH =- 2880 kJ mol^(-1)`
Enthalpy consumed in muscular work `=(2880kJ) ((25)/(100))`
`=720 kJ`
Amount of glucose in `120g = ((120g)/(180 g mol^(-1)))`
`= 0.6667 mol`
Enthalpy avaiable for muscular work from `120g` of glucose `=(720 kJ mol^(-1)) (0.667 mol) = 480 kJ`
Distance to which a person can move `= ((1km)/(100 kJ)) (480 kJ) = 4.80 km`