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Chapter 15: Problem 34

The dissociation of molecular iodine into iodine atoms is represented as $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ At \(1000 \mathrm{~K}\), the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(I_{2}\) in a 2.30 - \(L\) flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?

Short Answer

Expert verified
To solve the exercise, start by calculating the initial conditions and the changes in concentration as the reaction proceeds. Next, write the expression for the equilibrium constant Kc and substitute the formulas from the second step. You will get a quadratic equation which must be solved for 'x'. Once you have the value for 'x', substitute it back into the formulas from the second step to get the equilibrium concentrations of I2 and I.

Step by step solution

01

Initial Conditions

We know that initially, there are 0.0456 moles of I2, and the volume of the flask is 2.30L. As the reaction has not yet started, the initial concentration of I2 is \([I_2] = \frac{0.0456 \, moles}{2.30 \, L} = 0.0198 \, M\) and the concentration of I is 0 \( [I] = 0 \, M\) as it has not yet formed.
02

Changes in Concentration

As the reaction proceeds, the I2 will dissociate to give 2 molecules of I for each molecule of I2 that reacts. We can express this in terms of a change in concentration, represented as 'x', where \([I_2] = 0.0198 - x \, M\) and \([I] = 2x \, M\). 'x' represents the molar concentration of I2 that has dissociated.
03

Expression of \(K_c\)

The equilibrium constant, \(K_c\), is described by the expression \(K_c = \frac{[I]^2}{[I_2]}\). Substituting the expressions we found in Step 2, we get \(3.80 * 10^{-5} = \frac{(2x)^2}{0.0198 - x}\).
04

Solving for 'x'

This is a quadratic equation. Multiplying through by \(0.0198 - x\) and simplifying, we get \(4x^2 - 0.0198x - 3.80 * 10^{-5} = 0\). We then solve this for 'x'.
05

Find the Equilibrium Concentrations

The equilibrium concentrations are then given by substituting the value of 'x' found in Step 4 back into \([I_2] = 0.0198 - x \, M\) and \([I] = 2x \, M\) from Step 2.

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Most popular questions from this chapter

A 2.50 -mol quantity of \(\mathrm{NOCl}\) was initially placed in a 1.50-L reaction chamber at \(400^{\circ} \mathrm{C}\). After equilibrium was established, it was found that 28.0 percent of the NOCl had dissociated: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the equilibrium constant \(K_{c}\) for the reaction.

Heating solid sodium bicarbonate in a closed vessel established this equilibrium: \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system, (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system, (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.

Consider the potential energy diagrams for two types of reactions \(\mathrm{A} \rightleftharpoons \mathrm{B}\). In each case, answer the following questions for the system at equilibrium. (a) How would a catalyst affect the forward and reverse rates of the reaction? (b) How would a catalyst affect the energies of the reactant and product? (c) How would an increase in temperature affect the equilibrium constant? (d) If the only effect of a catalyst is to lower the activation energies for the forward and reverse reactions, show that the equilibrium constant remains unchanged if a catalyst is added to the reacting mixture.

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are: \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.491 \mathrm{M}\). The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(4.63 \times\) \(10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color.]

The vapor pressure of mercury is \(0.0020 \mathrm{mmHg}\) at \(26^{\circ} \mathrm{C}\). (a) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the process \(\mathrm{Hg}(l) \rightleftharpoons \mathrm{Hg}(g) .\) (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring \(6.1 \mathrm{~m}\) long, \(5.3 \mathrm{~m}\) wide, and \(3.1 \mathrm{~m}\) high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor in \(\mathrm{mg} / \mathrm{m}^{3}\). Does this concentration exceed the safety limit of \(0.050 \mathrm{mg} / \mathrm{m}^{3} ?\) (Ignore the volume of furniture and other objects in the laboratory.)
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