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Chapter 23: Problem 47

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

Short Answer

Expert verified
Answer: The equilibrium constant (K) for the photosynthesis reaction at 25°C is approximately 2.10 x 10⁻⁵⁰¹.

Step by step solution

01

Convert the temperature to Kelvin

To convert the temperature from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. $$ T_{Kelvin} = T_{Celsius}+273.15 $$ Plugging in the given temperature of 25°C, we get: $$ T_{Kelvin} = 25 + 273.15 = 298.15 \mathrm{~K} $$
02

Calculate the equilibrium constant \(K\) using the Gibbs free energy equation

Now that we have the temperature in Kelvin and we are given the standard Gibbs free energy change (\(\Delta G^{\circ}\)), we can use the equation: $$ \Delta G^{\circ} = -RT \ln K $$ to find \(K\). Rearranging the equation to solve for \(K\), we get: $$ K = e^{(-\Delta G^{\circ}/RT)} $$ Substitute the known values for \(\Delta G^{\circ} = 2.87 \times 10^{3}\mathrm{~kJ}\), \(R = 8.314 \mathrm{~J / mol \cdot K}\) (note the conversion of R from J to kJ), and \(T = 298.15\mathrm{~K}\) into the equation: $$ K = e^{(- (2.87 \times 10^{3}\mathrm{~kJ}) / (8.314 \times 10^{-3}\mathrm{~kJ / mol\cdot K}\times 298.15\mathrm{~K}))} $$
03

Solve for K

Now, we need to evaluate the expression for \(K\) using the values we have substituted: $$ K = e^{(- (2.87 \times 10^{3}) / (2.479\mathrm{~kJ/mol}))} $$ $$ K = e^{-1151.26} $$ $$ K \approx 2.10 \times 10^{-501} $$ Thus, the equilibrium constant (\(K\)) for the photosynthesis reaction at 25°C is approximately \(2.10 \times 10^{-501}\).

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Most popular questions from this chapter

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