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Chapter 13: Problem 49

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 in a \(2.0-\) L container 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} \mathrm{M} .\( Calculate the moles of \)\mathrm{O}_{2}(g)$ present under these conditions.

Short Answer

Expert verified
Under these conditions, there are approximately \(1.6 \times 10^{-2}\) moles of \(\mathrm{O}_2(g)\) present at equilibrium in the 2.0 L container.

Step by step solution

01

Write the expression for the equilibrium constant, K

Using the given reaction, the expression for K can be written as: \[K = \frac{[\mathrm{H}_2]^2[\mathrm{O}_2]}{[\mathrm{H_2O}]^2}\]
02

Plug in the given values

We are given \(K = 2.4 \times 10^{-3}\), $\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} \mathrm{M}$. Plug these values into the equilibrium constant expression: \[ 2.4 \times 10^{-3} = \frac{(1.9 \times 10^{-2})^2[\mathrm{O}_2]}{(1.1 \times 10^{-1})^2} \]
03

Solve for the concentration of O₂

Now we need to find the value of \([\mathrm{O}_2]\) by isolating it in the equation. \[ [\mathrm{O}_2]= \frac{2.4 \times 10^{-3}(1.1 \times 10^{-1})^2}{(1.9 \times 10^{-2})^2} \] Calculate the value: \[ [\mathrm{O}_2]\approx 7.8 \times 10^{-3} \mathrm{M} \]
04

Convert concentration to moles

Since the container has a volume of 2 L, we can find the moles of \(\mathrm{O}_2(g)\) using the calculated concentration: Moles of $\mathrm{O}_2 = [\mathrm{O}_2] \times V\] \[ = (7.8 \times 10^{-3}\mathrm{M}) \times (2\:\mathrm{L}) \] Calculate the moles of \(\mathrm{O}_2(g)\): \[ \text{Moles of}\:\mathrm{O}_2\approx 1.6 \times 10^{-2}\:\text{moles} \] Under these conditions, there are approximately \(1.6 \times 10^{-2}\) moles of \(\mathrm{O}_2(g)\) present at equilibrium in the 2.0 L container.

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Most popular questions from this chapter

Consider the decomposition of the compound $\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}$ as follows: $$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$ When a 5.63 -g sample of pure $\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g)\( was sealed into an otherwise empty \)2.50-\mathrm{L}$ flask and heated to \(200 .^{\circ} \mathrm{C},\) the pres- sure in the flask gradually rose to 1.63 \(\mathrm{atm}\) and remained at that value. Calculate \(K\) for this reaction.

For the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ at \(600 . \mathrm{K}\) , the equilibrium constant, \(K_{\mathrm{p}},\) is \(11.5 .\) Suppose that 2.450 \(\mathrm{g} \mathrm{PCl}_{5}\) is placed in an evacuated 500 -mL bulb, which is then heated to \(600 . \mathrm{K}\) . a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of PCl_ at equilibrium?

At \(25^{\circ} \mathrm{C},\) gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that 12.5\(\%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is 0.900 atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 -L vessel, calculate the equilibrium concentrations of all species.

An equilibrium mixture contains 0.60 g solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of 2.60 atm and 2.89 atm, respectively. Calculate the value of \(K_{\mathrm{p}}\) for the reaction $\mathrm{C}(s)+\mathrm{CO}_{2}(g) \Longrightarrow 2 \mathrm{CO}(g)$
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