It is observed that functioning of neuronal circuits in our brain is driven by the energy available from combustion of glucose. Calculate the amount of glucose (in gm) which should be burnt per hour to produce sufficient energy for brain, which operates at $\frac{128}{9}$ watts.
Given : $| Δ H^{0} |$ 400 K combustion of glucose = 3000 kJ/mole
$| Δ H^{0} |$ 400 K combustion of glucose = 180 kJ/mole K

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Solution

The correct option is D 3
$C_{6} H_{12} O_{6} (S) + 60_{2} (g) \to 6 C O_{2} (g) + 6 H_{2} O (g)$
$Δ G^{0} = Δ H^{0} - T Δ S^{0}$
(useful energy) $= - 3000 - 400 \times 0.18$
$= - 3072 k J / m o l e$
Energy required per hour $= \frac{128}{9} \times 60 \times 60 J$
$∴$ Glucose needed per hour $=$
$\frac{\frac{128}{9} \times 60 \times 60}{3072 \times 1000} m o l e = \frac{\frac{128}{9} \times 60 \times 60}{3072 \times 1000} \times 180 g = 3 g$

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The correct option is D 3C6H12O6(S)+602(g)→6CO2(g)+6H2O(g)ΔG0=ΔH0−TΔS0(useful energy)=−3000−400×0.18=−3072kJ/moleEnergy required per hour =1289×60×60J∴ Glucose needed per hour =1289×60×603072×1000mole=1289×60×603072×1000×180g=3g