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Chapter 13: Problem 54

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following cases. a. \(1.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) and \(2.0 \mathrm{~g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a \(1.0-\mathrm{L}\) flask. b. \(1.0\) mole of pure \(\mathrm{HOCl}\) is placed in a \(2.0-\mathrm{L}\) flask.

Short Answer

Expert verified
For case a, the equilibrium concentrations are [H2O]= 0.0439 M, [Cl2O]= 0.0281 M, and [HOCl]= 0.0232 M. For case b, the equilibrium concentrations are [H2O]= 0.229 M, [Cl2O]= 0.229 M, and [HOCl]= 0.042 M.

Step by step solution

01

Calculate initial concentrations

Convert the given masses of H2O and Cl2O to moles and then calculate their initial concentrations in a 1.0 L flask. H2O: 1.0 g / (18.02 g/mol) = 0.0555 mol Cl2O: 2.0 g / (50.45 g/mol) = 0.0397 mol Initial concentrations: [H2O] = 0.0555 mol / 1.0 L = 0.0555 M [Cl2O] = 0.0397 mol / 1.0 L = 0.0397 M [HOCl] = 0 M since it is not initially present.
02

Set up an ICE table

| | H2O | Cl2O | 2 HOCl | |-------|-----|-------|---------| | Initial | 0.0555 | 0.0397 | 0 | | Change | -x | -x | +2x | | Equilibrium | 0.0555-x | 0.0397-x | 2x |
03

Write the equilibrium constant expression

Use the given equilibrium constant K and substitute the equilibrium concentrations from the ICE table. \[K = \frac{[\mathrm{HOCl}]^2}{[\mathrm{H}_2 \mathrm{O}][\mathrm{Cl}_2 \mathrm{O}]} = \frac{(2x)^2}{(0.0555 - x)(0.0397 - x)} = 0.090\]
04

Solve the equation for x

Solve the equation for x either algebraically or using a numerical method like the quadratic formula. To simplify the math, we will assume that x is small so (0.0555 - x) ≈ 0.0555 and (0.0397 - x) ≈ 0.0397. We can revisit this assumption if necessary. \[0.090 = \frac{4x^2}{(0.0555)(0.0397)}\] Solving for x, we get x = 0.0116.
05

Calculate the equilibrium concentrations

Use the value of x to find the equilibrium concentrations of all species. [H2O] = 0.0555 - x = 0.0555 - 0.0116 = 0.0439 M [Cl2O] = 0.0397 - x = 0.0397 - 0.0116 = 0.0281 M [HOCl] = 2x = 2(0.0116) = 0.0232 M The equilibrium concentrations are [H2O]= 0.0439 M, [Cl2O]= 0.0281 M, and [HOCl]= 0.0232 M. #Case b#
06

Calculate initial concentrations

Calculate the initial concentration of HOCl in a 2.0 L flask. [HOCl] = 1.0 mol / 2.0 L = 0.5 M [H2O] = 0 M and [Cl2O] = 0 M since they are not initially present.
07

Set up an ICE table

| | H2O | Cl2O | 2 HOCl | |-------|-----|-------|---------| | Initial | 0 | 0 | 0.5 | | Change | +x | +x | -2x | | Equilibrium | x | x | 0.5-2x |
08

Write the equilibrium constant expression

Use the equilibrium constant K and substitute the equilibrium concentrations from the ICE table. \[K = \frac{[\mathrm{HOCl}]^2}{[\mathrm{H}_2 \mathrm{O}][\mathrm{Cl}_2 \mathrm{O}]} = \frac{(0.5 - 2x)^2}{x^2} = 0.090\]
09

Solve the equation for x

Solve the equation for x either algebraically or using a numerical method. \[0.090 = \frac{(0.5 - 2x)^2}{x^2}\] Solving for x, we get x = 0.229.
10

Calculate the equilibrium concentrations

Use the value of x to find the equilibrium concentrations of all species. [H2O] = x = 0.229 M [Cl2O] = x = 0.229 M [HOCl] = 0.5 - 2x = 0.5 - 2(0.229) = 0.042 M The equilibrium concentrations are [H2O]= 0.229 M, [Cl2O]= 0.229 M, and [HOCl]= 0.042 M.

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Most popular questions from this chapter

For the reaction $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3}\) \(M,\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{~N}_{2}(g)\right]=5.3 \times 10^{-2} M\), and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

Consider the reaction $$ \mathrm{P}_{4}(g) \longrightarrow 2 \mathrm{P}_{2}(g) $$ where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{~K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{~K}\), the equilibrium mixture of \(\mathrm{P}_{4}(\mathrm{~g})\) and \(\mathrm{P}_{2}(g)\) has a total pressure of \(1.00 \mathrm{~atm} .\) Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

In which direction will the position of the equilibrium $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\mathrm{HI}(g)\) is removed. d. In a rigid reaction container, some \(\operatorname{Ar}(g)\) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).

What will happen to the number of moles of \(\mathrm{SO}_{3}\) in equilibrium with \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) in the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ in each of the following cases? a. Oxygen gas is added. b. The pressure is increased by decreasing the volume of the reaction container. c. In a rigid reaction container, the pressure is increased by adding argon gas. d. The temperature is decreased (the reaction is endothermic). e. Gaseous sulfur dioxide is removed.

The equilibrium constant is \(0.0900\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0-L flask contains \(1.0\) mole of \(\mathrm{HOCl}, 0.10\) mole of \(\mathrm{Cl}_{2} \mathrm{O}\), and \(0.10\) mole of \(\mathrm{H}_{2} \mathrm{O}\). b. A \(2.0\) - \(\mathrm{L}\) flask contains \(0.084\) mole of \(\mathrm{HOCl}, 0.080\) mole of \(\mathrm{Cl}_{2} \mathrm{O}\), and \(0.98 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}\). c. A 3.0-L flask contains \(0.25\) mole of HOCl, \(0.0010\) mole of \(\mathrm{Cl}_{2} \mathrm{O}\), and \(0.56 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}\).
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