The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 20. kg of sucrose (C12H22O11) per hour per hectare (1 ha = the reaction 10,000 m²). Assuming that sucrose is produced by 12CO2 (g) + 11H20(1) · → C12 H22 O11 (s) + 1202(g) AH = 5640 kJ calculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis in this field. Percent efficiency =
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
![The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 20. kg of sucrose (C₁₂H₂₂O₁₁) per hour per hectare (1 ha = 10,000 m²). Assuming that sucrose is produced by the reaction
\[ 12\text{CO}_2(g) + 11\text{H}_2\text{O}(l) \rightarrow \text{C}_{12}\text{H}_{22}\text{O}_{11}(s) + 12\text{O}_2(g) \quad \Delta H = 5640 \text{ kJ} \]
Calculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis in this field.
Percent efficiency = [ ] %
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