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Chapter 14: Problem 17

The process of photosynthesis in plants converts carbon dioxide and water to glucose and oxygen: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \quad \Delta E_{\mathrm{rxn}}=2801 \mathrm{~kJ}\) (a) Write the equilibrium constant expression for this conversion. (b) How would the equilibrium be affected if \(\mathrm{CO}_{2}(g)\) were added? (c) How would the equilibrium be affected if \(\mathrm{H}_{2} \mathrm{O}(l)\) were added? (d) How would the equilibrium be affected if the reaction vessel were warmed? (e) How would the equilibrium be affected if a catalyst were added?

Short Answer

Expert verified
\( K = \frac{[\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}][\mathrm{O}_{2}]^{6}}{[\mathrm{CO}_{2}]^{6}[\mathrm{H}_{2}\mathrm{O}]^{6}} \) (a) The equilibrium constant expression is K as stated above. (b) Adding CO₂ shifts the equilibrium to the right. (c) Adding H₂O shifts the equilibrium to the right. (d) Warming the reaction vessel shifts the equilibrium to the right. (e) Adding a catalyst does not affect the position of the equilibrium.

Step by step solution

01

(a) Write the equilibrium constant expression

To write the expression for the equilibrium constant K, we need to consider the concentrations of products divided by the concentrations of reactants, each raised to the power of their stoichiometric coefficients. In this case, we have: \( K = \frac{[\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}][\mathrm{O}_{2}]^{6}}{[\mathrm{CO}_{2}]^{6}[\mathrm{H}_{2}\mathrm{O}]^{6}} \)
02

(b) Effect of adding CO₂

According to Le Chatelier's principle, when stress is applied to a system at equilibrium, the system will shift to minimize the stress. In this case, if we add CO₂, the system will shift towards the products side to consume the added CO₂. Thus, the equilibrium would shift to the right.
03

(c) Effect of adding H₂O

Similar to the CO₂ case, when we add H₂O, the system will shift to minimize the stress. In this case, the system will shift towards the products side to consume the added H₂O. Thus, the equilibrium would again shift to the right.
04

(d) Effect of warming the reaction vessel

Since the reaction is endothermic (\( \Delta E_{\mathrm{rxn}} > 0 \)), increasing the temperature will favour the side of the reaction that absorbs heat. In this case, this is the side of the products, meaning that increasing the temperature will shift the equilibrium towards the right.
05

(e) Effect of adding a catalyst

A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction. Adding a catalyst to the system will increase the rate of both the forward and reverse reactions, without affecting their equilibrium concentrations. Therefore, the equilibrium constant K and the position of the equilibrium will remain unchanged.

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Most popular questions from this chapter

When the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) is run at \(2000^{\circ} \mathrm{C}\), appreciable amounts of reactants and product are present at equilibrium. (a) A sealed 2.00-L container at \(2000{ }^{\circ} \mathrm{C}\) is filled with \(1.00\) mole of \(\mathrm{NO}(g)\) and nothing else. At that moment, which reaction is faster, forward or reverse? Justify your answer. (b) At equilibrium, the concentration of \(\mathrm{NO}(g)\) is \(0.0683 \mathrm{M}\) and the concentration of \(\mathrm{N}_{2}(g)\) is \(0.2159 \mathrm{M}\). What is the value of \(K_{\mathrm{eq}}\) at \(2000^{\circ} \mathrm{C} ?\)

What effect does a catalyst have on: (a) The position of equilibrium for a reaction? (b) The value of the equilibrium constant? (c) The ratio of \(k_{\mathrm{f}} / k_{\mathrm{r}} ?\) 14.86 What does a catalyst do to the time it takes for a reaction to reach equilibrium? Explain how it does this.

\(K_{\text {eq }}=3.9 \times 10^{-11}\) for the dissolution of calcium fluoride in water: \(\mathrm{CaF}_{2}(s) \rightleftarrows \mathrm{Ca}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)\) (a) What is another name for \(K_{\mathrm{eq}}\) for this reaction? (b) If the equilibrium calcium ion concentration in a saturated aqueous solution of calcium fluoride is \(3.3 \times 10^{-4} \mathrm{M}\), what is the equilibrium fluoride ion concentration? (c) Which is larger, the rate constant for the forward reaction or the rate constant for the reverse reaction? (d) Which is larger, \(E_{a}\) for the forward reaction or \(E_{\mathrm{a}}\) for the reverse reaction? (e) Which is larger, the rate of the forward reaction or the rate of the reverse reaction? (f) For lithium carbonate, \(K_{\mathrm{sp}}=0.0011\). Write the balanced chemical equation and the equilibrium expression for the dissolution of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) in water. (g) Which is more soluble in water, calcium fluoride or lithium carbonate?

Sparingly soluble \(\mathrm{PbCl}_{2}\) dissolves in water to yield an equilibrium \(\mathrm{Pb}^{2+}(a q)\) concentration of \(0.039 \mathrm{M}\). (a) Write the balanced equilibrium equation for \(\mathrm{PbCl}_{2}(\) s) dissolving in water. (b) Write the \(K_{\text {sp }}\) expression for \(\mathrm{PbCl}_{2}\). (c) What is the equilibrium concentration of chloride ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{PbCl}_{2}\) (show your calculation).

How can you quickly determine the saturation solubility of a sparingly soluble \(1: 1\) salt at \(25^{\circ} \mathrm{C}\) ?
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