Standard molar enthalpy of combustion of glucose is $- 2880 k J$ . If only $25$ % of energy is available for muscular work and $1.0 k m$ walk consume $90 k J$ of energy, what is the maximum distance (in km) a person can walk after eating $90 g$ of glucose.

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Solution

90 gm glucose means 0.5 mole of it will give energy of 1440kJ and 25% of it is used in walking so energy used $= 0.25 \times 1440 = 360 k J$
So, distance travel by a person is $x \times 90 = 360; \Rightarrow x = 4 k m$

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Standard molar enthalpy of combustion of glucose is −2880kJ. If only 25% of energy is available for muscular work and 1.0km walk consume 90kJ of energy, what is the maximum distance (in km) a person can walk after eating 90g of glucose.

90 gm glucose means 0.5 mole of it will give energy of 1440kJ and 25% of it is used in walking so energy used =0.25×1440=360 kJSo, distance travel by a person is x×90=360;⇒x=4km

Standard molar enthalpy of combustion of glucose is $- 2880 k J$ . If only $25$ % of energy is available for muscular work and $1.0 k m$ walk consume $90 k J$ of energy, what is the maximum distance (in km) a person can walk after eating $90 g$ of glucose.

90 gm glucose means 0.5 mole of it will give energy of 1440kJ and 25% of it is used in walking so energy used $= 0.25 \times 1440 = 360 k J$
So, distance travel by a person is $x \times 90 = 360; \Rightarrow x = 4 k m$