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Chapter 15: Problem 47

Consider the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ What would be the effect on the position of equilibrium of (a) increasing the total pressure on the system by decreasing its volume, (b) adding \(I_{2}\) to the reaction mixture, (c) decreasing the temperature?

Short Answer

Expert verified
According to Le Chatelier's Principle, (a) increasing pressure will cause the equilibrium to shift towards the side with fewer gaseous molecules—in this case, the formation of \(I_{2}\), (b) adding \(I_{2}\) will make the equilibrium shift to the left, favoring dissociation into two \(I\) atoms, and (c) lowering the temperature will shift the equilibrium to the right, producing more \(I_{2}\) since the reaction is endothermic.

Step by step solution

01

Effect of Increasing Pressure

According to Le Chatelier's Principle, if pressure is increased, the equilibrium shifts in the direction where fewer gaseous molecules exist, to reduce the pressure. In the equation \(2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g)\), the right-hand side (1 mol of \(I_{2}\)) has fewer molecules than the left-hand side (2 mol of \(I\)). Therefore, when the pressure is increased, the equilibrium will shift to the right to favor the formation of more \(I_{2}\).
02

Effect of Additional Iodine \(I_{2}\)

When more \(I_{2}\) is added to the mixture, the increased concentration of \(I_{2}\) would cause a shift to the side with less \(I_{2}\) to establish a new equilibrium according to Le Chatelier's principle. As a result, the equilibrium will shift to the left, favoring the dissociation into two \(I\) molecules.
03

Effect of Decreasing Temperature

This reaction is endothermic (absorbs heat) because the dissociation of iodine into two separate atoms requires energy. So according to Le Chatelier's principle, if the temperature decreases (heat is removed), the equilibrium will shift in the direction that produces heat. That would be the left hand side, as 2I is converted to \(I_{2}\). So the equilibrium will shift to the right, yielding a significant increase in \(I_{2}\).

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Most popular questions from this chapter

Iodine is sparingly soluble in water but much more so in carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\). The equilibrium constant, also called the partition coefficient, for the distribution of \(\mathrm{I}_{2}\) between these two phases $$ \mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) $$ is 83 at \(20^{\circ} \mathrm{C}\). (a) A student adds \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\) to \(0.200 \mathrm{~L}\) of an aqueous solution containing \(0.032 \mathrm{~g}\) \(\mathrm{I}_{2} .\) The mixture is shaken and the two phases are then allowed to separate. Calculate the fraction of \(\mathrm{I}_{2}\) remaining in the aqueous phase. (b) The student now repeats the extraction of \(I_{2}\) with another 0.030 \(\mathrm{L}\) of \(\mathrm{CCl}_{4}\). Calculate the fraction of the \(\mathrm{I}_{2}\) from the original solution that remains in the aqueous phase. (c) Compare the result in (b) with a single extraction using \(0.060 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\). Comment on the difference.

The equilibrium constant \(K_{\mathrm{c}}\) for the decomposition of phosgene, \(\mathrm{COCl}_{2}\), is \(4.63 \times 10^{-3}\) at \(527^{\circ} \mathrm{C}\) : $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.760 atm.

Pure NOCl gas was heated at \(240^{\circ} \mathrm{C}\) in a 1.00 - \(\mathrm{L}\) container. At equilibrium the total pressure was 1.00 atm and the NOCl pressure was 0.64 atm. $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate the partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) in the system. (b) Calculate the equilibrium constant \(K_{P}\).

Write the equilibrium constant expressions for \(K_{\mathrm{c}}\) and \(K_{P}\), if applicable, for these reactions: (a) \(2 \mathrm{NO}_{2}(g)+7 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q)\)

A reaction vessel contains \(\mathrm{NH}_{3}, \mathrm{~N}_{2},\) and \(\mathrm{H}_{2}\) at equilibrium at a certain temperature. The equilibrium concentrations are \(\left[\mathrm{NH}_{3}\right]=0.25 \mathrm{M},\left[\mathrm{N}_{2}\right]=0.11 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=1.91 \mathrm{M}\). Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the synthesis of ammonia if the reaction is represented as (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\)
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