The formation of glucose from water and carbon dioxide is one of the more important chemical reactions in the world. Plants perform this reaction through the process of photosynthesis, creating the base of the food chain: $$ 6 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ At a particular temperature, the following equilibrium concentrations were found: \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=7.91 \times 10^{-2} M,\left[\mathrm{CO}_{2}(g)\right]=\) \(9.3 \times 10^{-1} M\), and \(\left[\mathrm{O}_{2}(g)\right]=2.4 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

Step by step solution

01

Write the expression for the equilibrium constant (K)

The equilibrium constant (K) is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. Using the given equation for photosynthesis, we can write the expression for K as: \[K = \frac{[\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}]^1[\mathrm{O}_{2}]^6}{[\mathrm{H}_{2}\mathrm{O}]^6 [\mathrm{CO}_{2}]^6}\]

02

Plug in the given equilibrium concentrations

Now, substitute the equilibrium concentrations given in the problem (\([\mathrm{H}_{2}\mathrm{O}] = 7.91\times10^{-2} M\), \([\mathrm{CO}_{2}] = 9.3\times10^{-1} M\), and \([\mathrm{O}_{2}] = 2.4\times10^{-3} M\)) into the expression for K. Note that we are not given the concentration of glucose, so we can assume it remains constant and does not affect the calculation of K. \[K = \frac{[\mathrm{O}_{2}]^6}{[\mathrm{H}_{2}\mathrm{O}]^6 [\mathrm{CO}_{2}]^6} = \frac{(2.4\times10^{-3})^6}{(7.91\times10^{-2})^6 (9.3\times10^{-1})^6}\]

03

Calculate K

Now, we can calculate the equilibrium constant using the substituted values: \[ K = \frac{(2.4\times10^{-3})^6}{(7.91\times10^{-2})^6 (9.3\times10^{-1})^6} \approx 2.74\times10^{-24} \] Thus, the value of K for the photosynthesis reaction at this temperature is approximately \(2.74\times10^{-24}\).

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