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Chapter 13: Problem 43

For the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ \(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} M\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} M\) What is the concentration of \(\mathrm{O}_{2}(g)\) under these conditions?

Short Answer

Expert verified
The concentration of \(\text{O}_{2}(g)\) under the given equilibrium conditions is approximately \(8.04 \times 10^{-2} M\).

Step by step solution

01

Write the equilibrium constant expression

For the given chemical equilibrium reaction, we can write the equilibrium constant expression as: \[K = \frac{[\text{H}_{2}]^2[\text{O}_{2}]}{[\text{H}_{2}\text{O}]^2}\]
02

Plug in the given values and solve for [O₂(g)]

We are given \(K = 2.4 \times 10^{-3}\), \([\text{H}_{2}\text{O}(g)] = 1.1 \times 10^{-1} M\), and \([\text{H}_{2}(g)] = 1.9 \times 10^{-2} M\). We can plug these values into the equilibrium constant expression and solve for \([\text{O}_{2}(g)]\): \[ 2.4 \times 10^{-3} = \frac{(1.9 \times 10^{-2})^2 [\text{O}_{2}]}{(1.1 \times 10^{-1})^2} \] Now, we will solve for \([\text{O}_{2}(g)]\): \[ [\text{O}_{2}] = \frac{2.4 \times 10^{-3} (1.1 \times 10^{-1})^2}{(1.9 \times 10^{-2})^2} \]
03

Calculate the concentration of O₂(g)

Perform the calculation: \[ [\text{O}_{2}] = \frac{2.4 \times 10^{-3} (1.21 \times 10^{-2})}{(3.61 \times 10^{-4})} \] \[ [\text{O}_{2}] = \frac{2.904 \times 10^{-5}}{3.61 \times 10^{-4}} \] \[ [\text{O}_{2}] \approx 8.04 \times 10^{-2} M \] The concentration of \(\text{O}_{2}(g)\) under the given equilibrium conditions is approximately \(8.04 \times 10^{-2} M\).

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Most popular questions from this chapter

As we have attempted to lessen our dependence on fossil fuels, the demand for biofuels, such as ethanol, which is produced by the fermentation of the sugars found in corn, has increased. Using Le Châtelier's principle, predict which way the equilibrium will shift during the fermentation of sugar for each of the following changes. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(a q) $$ a. when the concentration of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is increased b. when the concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is decreased c. when \(\mathrm{CO}_{2}\) gas is added to the solution d. when the volume of water in the solution is doubled

Trans fats, or partially hydrogenated cooking oils, result from efforts of food manufacturers to increase the shelf life of many different forms of cooking oils. The free-flowing oils are treated with hydrogen, a catalyst, and high heat to change the structure of the oil, raising the melting point. The now solid oil is much more stable toward spoilage. A serious drawback to this process has been the realization that the trans form of the fat poses serious health risks when compared with the naturally occurring cis forms of the oils. The part of the hydrogenation process that results in the formation of trans fats can be summarized as follows: $$ \text { cis fat }(a q)+\mathrm{H}_{2}(a q) \rightleftharpoons \operatorname{trans} \operatorname{fat}(a q) $$ A typical value of \(K\) for this type of reaction is about \(5.0 .\) If we start with \(0.10 \mathrm{~mol} / \mathrm{L}\) each of cis fat and \(\mathrm{H}_{2}\), calculate the equilibrium concentrations.

Consider the reaction $$ \mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q) $$ How will the equilibrium position shift if a. water is added, doubling the volume? b. \(\mathrm{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? \(\left[\mathrm{Fe}(\mathrm{OH})_{3}\right.\) is insoluble.] d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added?

At a particular temperature, \(K=4.0 \times 10^{-7}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ In an experiment, \(1.0 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}\) is placed in a 10.0-L vessel. Calculate the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) when this reaction reaches equilibrium.

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ If \(2.0 \mathrm{~mol} \mathrm{NO}\) and \(1.0 \mathrm{~mol} \mathrm{Cl}_{2}\) are placed into a \(1.0\) - \(\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.
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