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Chapter 13: Problem 60

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 -L flask. b. 1.0 mole of pure HOCl is placed in a 2.0 \(\mathrm{L}\) flask.

Short Answer

Expert verified
For case a, the equilibrium concentrations are: \[ [\text{H}_{2}\text{O}] = 0.0381 \, \text{M} \], \[ [\text{Cl}_{2}\text{O}] = 0.0231 \, \text{M} \], and \[ [\text{HOCl}] = 0.0338 \, \text{M} \]. For case b, the equilibrium concentrations are: \[ [\text{H}_{2}\text{O}] = 0.212 \, \text{M} \], \[ [\text{Cl}_{2}\text{O}] = 0.212\, \text{M} \], and \[ [\text{HOCl}] = 0.076 \, \text{M} \].

Step by step solution

01

Calculate moles of H2O and Cl2O at the beginning

Use the given masses and molar masses of H2O (18.02 g/mol) and Cl2O (50.45 g/mol) to calculate the moles of each present at the start: Moles of H2O = \( \frac{1.0 \, \text{g}}{18.02 \, \text{g/mol}} = 0.055 \, \text{mol} \) Moles of Cl2O = \( \frac{2.0 \, \text{g}}{50.45 \, \text{g/mol}} = 0.040 \, \text{mol} \)
02

Set up the initial and equilibrium concentrations

Create an ICE (initial, change, equilibrium) table to find the equilibrium concentrations. Initially, the concentration of HOCl is zero because it hasn't been formed yet. | | H2O | Cl2O | HOCl | |-----|-----|------|-------| | I | 0.055 | 0.040 | 0 | | C | -x | -x | +2x | | E | 0.055-x | 0.040-x | 2x |
03

Set up the equilibrium expression and solve for x

Write the equilibrium expression for K and plug in the equilibrium concentrations from the ICE table: \( K = \frac{[\text{HOCl}]^2}{[\text{H}_{2}\text{O}][\text{Cl}_{2}\text{O}]} \) Substitute with the values from the ICE table: \( 0.090 = \frac{(2x)^2}{(0.055-x)(0.040-x)} \) Now, solve the equation for x. Since K is small, we can estimate that x is small compared to 0.055 and 0.040, so we can simplify the expression to: \( 0.090 = \frac{4x^2}{(0.055)(0.040)} \) Solve for x: \( x = 0.0169 \)
04

Calculate the equilibrium concentrations

Substitute the value of x back into the equilibrium values from the ICE table to find the concentrations at equilibrium: \[ [\text{H}_{2}\text{O}] = 0.055 - 0.0169 = 0.0381 \, \text{M} \] \[ [\text{Cl}_{2}\text{O}] = 0.040 - 0.0169 = 0.0231 \, \text{M} \] \[ [\text{HOCl}] = 2(0.0169) = 0.0338 \, \text{M} \] #Case b: Initial mixture of pure HOCl#
05

Calculate initial concentration of HOCl

Calculate the initial concentration of HOCl by dividing the moles by the volume: \[ [\text{HOCl}]_{i} = \frac{1.0 \, \text{mol}}{2.0 \, \text{L}} = 0.5 \, \text{M} \]
06

Set up the initial and equilibrium concentrations

Create an ICE table to find the equilibrium concentrations. Initially, the concentrations of H2O and Cl2O are zero because they haven't been formed yet. | | H2O | Cl2O | HOCl | |-----|-----|------|-------| | I | 0 | 0 | 0.5 | | C | +x | +x | -2x | | E | x | x | 0.5-2x |
07

Set up the equilibrium expression and solve for x

Write the equilibrium expression for K and plug in the equilibrium values from the ICE table: \( K = 0.090 = \frac{x^2}{(0.5-2x)} \) Now, solve the equation for x. Since K is small, we can estimate that 2x is small compared to 0.5, so the equation simplifies to: \( 0.090 = \frac{x^2}{0.5} \) Solve for x: \( x = 0.212 \)
08

Calculate the equilibrium concentrations

Substitute the value of x back into the equilibrium values from the ICE table to find the concentrations at equilibrium: \[ [\text{H}_{2}\text{O}] = x = 0.212 \, \text{M} \] \[ [\text{Cl}_{2}\text{O}] = x = 0.212\, \text{M} \] \[ [\text{HOCl}] = 0.5 - 2x = 0.5 - 2(0.212) = 0.076 \, \text{M} \]

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Most popular questions from this chapter

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ What is the partial pressure of NO in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of 0.80 and \(0.20 \mathrm{atm},\) respectively?

The reaction to prepare methanol from carbon monoxide and hydrogen $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ is exothermic. If you wanted to use this reaction to produce methanol commercially, would high or low temperatures favor a maximum yield? Explain.

Consider the reaction $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ Suppose the system is at equilibrium, and then an additional mole of \(\mathrm{N}_{2} \mathrm{O}(g)\) is injected into the system at constant temperature. Once the reaction reestablishes equilibrium, has the amount of \(\mathrm{N}_{2} \mathrm{O}\) increased or decreased from its original equilibrium amount? Explain. What happens to the value of the equilibrium constant with this change?

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 HOCl, 0.10 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.10 mole of \)\mathrm{H}_{2} \mathrm{O}$ . b. A 2.0 -L flask contains 0.084 mole of HOCl, 0.080 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.98 mole of \)\mathrm{H}_{2} \mathrm{O}$ . c. A 3.0 - flask contains 0.25 mole of HOCl, 0.0010 mole of $\mathrm{Cl}_{2} \mathrm{O},\( and 0.56 mole of \)\mathrm{H}_{2} \mathrm{O}$ .

At a particular temperature, \(K=3.75\) for the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$ If all four gases had initial concentrations of \(0.800 M,\) calculate the equilibrium concentrations of the gases.
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