Question 10.3: Given the thermochemical equation for photosynthesis, 6H2O(l...
Given the thermochemical equation for photosynthesis,
6\text{H}_2\text{O}(l) + 6\text{CO}_2(\text{g}) → \text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(\text{g}) ∆H = +2803 \text{kJ/mol}
calculate the solar energy required to produce 75.0 \text{g} of \text{C}_6\text{H}_{12}\text{O}_6.
Strategy The thermochemical equation shows that for every mole of \text{C}_6\text{H}_{12}\text{O}_6 produced, 2803 \text{kJ} is absorbed. We need to find out how much energy is absorbed for the production of 75.0 \text{g} of \text{C}_6\text{H}_{12}\text{O}_6. We must first find out how many moles there are in 75.0 \text{g} of \text{C}_6\text{H}_{12}\text{O}_6.
Setup The molar mass of \text{C}_6\text{H}_{12}\text{O}_6 is 180.2 \text{g/mol}, so 75.0 \text{g} of \text{C}_6\text{H}_{12}\text{O}_6 is
75.0 \text{g} \text{C}_6\text{H}_{12}\text{O}_6 × \frac{1 \text{mol} \text{C}_6\text{H}_{12}\text{O}_6}{180.2 \text{g} \text{C}_6\text{H}_{12}\text{O}_6} = 0.416 \text{mol} \text{C}_6\text{H}_{12}\text{O}_6
We will multiply the thermochemical equation, including the enthalpy change, by 0.416, in order to write the equation in terms of the appropriate amount of \text{C}_6\text{H}_{12}\text{O}_6.
(0.416 \text{mol})[6\text{H}_2\text{O}(l) + 6\text{CO}_2(\text{g}) → \text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(\text{g})]
and (0.416 \text{mol)(∆H)} = (0.416 \text{mol})(2803 \text{kJ/mol}) gives
2.50 \text{H}_2\text{O}(l) + 2.50 \text{CO}_2(\text{g}) → 0.416 \text{C}_6\text{H}_{12}\text{O}_6(s) + 2.50 \text{O}_2(g) ∆H = + 1.17 × 10^3 \text{kJ}
Therefore, 1.17 × 10^3 \text{kJ} of energy in the form of sunlight is consumed in the production of 75.0 \text{g} of \text{C}_6\text{H}_{12}\text{O}_6. Note that the “per mole” units in ∆H are canceled when we multiply the thermochemical equation by the number of moles of \text{C}_6\text{H}_{12}\text{O}_6.