Question

Plants synthesize carbohydrates from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by the process of photosynthesis. For example, $$ 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) $$ \(\Delta G^{\circ}=2.87 \times 10^{3} \mathrm{~kJ}\) at \(\mathrm{pH} 7.0\) and \(25^{\circ} \mathrm{C}\). What is \(K\) for the reaction at \(25^{\circ} \mathrm{C}\) ?

Short answer

Answer: The equilibrium constant (K) for the photosynthesis reaction at 25°C is approximately 9.1 × 10⁻⁶.

Step-by-step solution

Convert temperature to Kelvin

The first step is to convert the given temperature of 25°C to Kelvin. To do this, add 273.15 to the Celsius temperature: $$T(K) = 25 + 273.15 = 298.15\,\mathrm{K}$$

Apply Gibbs free energy change and equilibrium constant relationship

Next, we will use the relationship between \(\Delta G^{\circ}\) and \(K\), which is given by the equation: $$\Delta G^{\circ}=-RT \ln K$$ We need to solve for \(K\). Rearrange the equation to isolate \(K\): $$K = e^{-\frac{\Delta G^{\circ}}{RT}}$$

Substitute the values and calculate \(K\)

Now, we can substitute the given values for \(\Delta G^{\circ}\), \(R\), and \(T\) in the equation to calculate \(K\). $$K = e^{-\frac{2.87 \times 10^{3} \mathrm{~kJ}}{(8.314 \times 10^{-3} \mathrm{~kJ/mol\cdot K})(298.15\,\mathrm{K})}}$$ First, calculate the value in the exponent: $$-\frac{2.87 \times 10^{3}}{(8.314 \times 10^{-3})(298.15)} = -11.6$$ Next, use the calculated exponent to find the equilibrium constant: $$K = e^{-11.6} \approx 9.1 \times 10^{-6}$$

Interpret the result

The equilibrium constant (\(K\)) for the photosynthesis reaction at 25°C is approximately \(9.1 \times 10^{-6}\). This small value indicates that the reaction does not strongly favor the formation of products at this temperature; instead, the reaction lies significantly towards the reactants.