Question

At the end of 2012, global population was about 7.0 billion people. What mass of glucose in kg would be needed to provide 1500 cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to CO2(g) and H2O(l) according to the following thermochemical equation:

C₆H₁₂O₆(s)+6O₂(g)⟶6CO₂(g)+6H₂O(l)

ΔH∘=−2803kJ

Answer 1

Given :

Number of persons = 7 billion = 7*10⁹ people

Consumption of glucose = 1500 cal/person/day

Total calories for 7*10⁹ people and 365 days (1 year) = 1500 cal/person/day * 7*10⁹ people*365 days = 3.8325*10¹⁵ cal

The conversion of cal into kJ is shown below:

1 Cal = 4.186* 10^-3 kJ

So,

Consumption of glucose = 1.6043*10¹³ kJ

Considering the given reaction,

2803 kJ of energy is released when 1 mole of glucose is metabolized.

1 kJ of energy is released when 1/2803 mole of glucose is metabolized.

So,

1.6043*10¹³ kJ of energy is released when (1/2803)*1.6043*10¹³ moles of glucose is metabolized.

Hence, moles of glucose = 5.7235*10⁹ moles

Molar mass of glucose = 180.156 g / mol

So,

Mass of glucose required = 5.7235*10⁹ moles * 180.156 g / mol = 1.0311*10¹² g

The conversion of g into kg is shown below:

1 g = 10^-3 kg

So,

Mass of glucose required = 1.0311*10⁹ kg