Question

The total surface area for the Asia consisting of forest, cultivated land, grass land, and desert is approximately 4.46 x 107 km2. Every year, the mass of carbon fixed by photosynthesis by vegetation covering this land surface according to reaction 6 CO2(g) + 6H2O (l)  C6H12O6(s) + 6O2(g) is about 455 x 103 kg km-2. Calculate the annual enthalpy change resulting from the photosynthetic carbon fixation over the land surface. Assume 1 bar and 298 K.

Answer 1

Explanation:

We calculate the amount of carbon fixed as follows.

          $455 \times 10^{3} kg/km^{2} \times 4.46 \times 10^{7} km^{2}$

            = $2029.3 \times 10^{10}$ kg

$\Delta H_{f}_{C_{6}H_{12}O_{6}}$ = -1273.1 kJ/mol

$\Delta H_{f}_{CO_{2}}$ = -393.5 kJ/mol

$\Delta H_{f}_{H_{2}O}$ = -285.8 kJ/mol

Hence, total $\Delta H$ reaction will be as follows.

             $\Delta H_{rxn} = (\Delta H_{f}_{C_{6}H_{12}O_{6}}) - 6(\Delta H_{f}_{CO_{2}}) - 6(\Delta H_{f}_{H_{2}O})$

                           = $-1273.1 kJ/mol - (6 \times 393.5 kJ/mol) - (6 \times 285.8 kJ/mol)$

                           = 2802.7 kJ/mol

Therefore, calculate the number of moles of fixed carbon as follows.

           $n_{c, fixed} = \frac{m_{c, fixed}}{\text{Molar mass of C in kg/mol}}$

                   = $\frac{4.55 \times 10^{5}kg km^{-2} year^{-1}}{12.01 \times 10^{-3} kg/mol}$

                   = $3.786 \times 10^{7} mol km^{-2} year^{-1}$

Thus, we can conclude that the annual enthalpy change resulting from the photosynthetic carbon fixation over the land surface is $3.786 \times 10^{7} mol km^{-2} year^{-1}$.