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Chapter 12: 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 HOCl is placed in a 2.0 -L flask.

Short Answer

Expert verified
In case a, the equilibrium concentrations are [H2O] = 0.0309 M, [Cl2O] = 0.0151 M, and [HOCl] = 0.0492 M. In case b, the equilibrium concentrations are [H2O] = [Cl2O] = 0.158 M, and [HOCl] = 0.184 M.

Step by step solution

01

Calculate moles of reactants

First, we need to convert the given masses of the reactants into moles using their molar masses. Molar mass of H2O = 18.015 g/mol Moles of H2O: \(\frac{1.0 \, g}{18.015 \, g/mol} = 0.0555 \, mol\) Molar mass of Cl2O = 50.45 g/mol Moles of Cl2O: \(\frac{2.0 \, g}{50.45 \, g/mol} = 0.0397 \, mol\)
02

Set up ICE table and plug in values

An ICE table allows us to keep track of the moles of each species at different stages of the reaction. ``` Initial Change Equilibrium H2O 0.0555 -x 0.0555 - x Cl2O 0.0397 -x 0.0397 - x HOCl 0 +2x 2x ```
03

Use equilibrium constant K to form an equation

Given K = 0.090. The formula for K: \(K = \frac{[HOCl]^2}{[H_2O][Cl_2O]}\) Plug in values from the equilibrium row of the ICE table to the K formula: 0.090 = \(\frac{(2x)^2}{(0.0555 - x)(0.0397 - x)}\)
04

Solve for x

Solve the above equation for the unknown variable x. x ~ 0.0246
05

Find concentrations at equilibrium

Now plug in the value of x into the equilibrium row of the ICE table to find concentrations: [H2O] = 0.0555 - 0.0246 = 0.0309 M [Cl2O] = 0.0397 - 0.0246 = 0.0151 M [HOCl] = 2 × 0.0246 = 0.0492 M These are the equilibrium concentrations for case a. Case b: b. 1.0 mole of pure HOCl is placed in a 2.0-L flask.
06

Set up ICE table and plug in values

Given the number of moles and the volume of the flask, we can directly write down the initial concentrations of each species: ``` Initial Change Equilibrium H2O 0 +x x Cl2O 0 +x x HOCl 0.50 -2x 0.50-2x ```
07

Use equilibrium constant K to form an equation

Given K = 0.090. The formula for K: \(K = \frac{[HOCl]^2}{[H_2O][Cl_2O]}\) Plug in values from the equilibrium row of the ICE table to the K formula: 0.090 = \(\frac{(0.50 - 2x)^2}{x^2}\)
08

Solve for x

Solve the above equation for the unknown variable x. x ~ 0.158
09

Find concentrations at equilibrium

Now plug in the value of x into the equilibrium row of the ICE table to find concentrations: [H2O] = [Cl2O] = 0.158 M [HOCl] = 0.50 - (2 × 0.158) = 0.184 M These are the equilibrium concentrations for case b.

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Most popular questions from this chapter

The equilibrium constant \(K_{\mathrm{p}}\) for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ at \(700^{\circ} \mathrm{C}\) is \(0.76 .\) Determine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of 1.20 atm at \(700^{\circ} \mathrm{C}\).

Consider the following reaction:$$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$.Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2},\) and \(\mathrm{CO}_{2}\) are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \(^{14} \mathrm{C},\) will \(^{14} \mathrm{C}\) be found only in CO molecules for an indefinite period of time? Explain.

Suppose a reaction has the equilibrium constant \(K=1.3 \times 10^{8} .\) What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?

Consider the decomposition equilibrium for dinitrogen pentoxide: $$2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2(g)+\mathrm{O}_{2}(g)$$.At a certain temperature and a total pressure of 1.00 atm, the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0,\) will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to 0.50\%? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of 10.0

Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$2 \mathrm{A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g)$$what is the value of \(K\) at the same temperature for the reaction $$ \mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{B}(g)$$.What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with 1.50 atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D},\) what is the mole fraction of B once equilibrium is reached?
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