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Chapter 13: Problem 11

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 \(\underline{\mathrm{a}}\) 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.

Short Answer

Expert verified
In the given chemical equilibrium, radioactive \( {}^{14}\mathrm{C} \) will not be found only in CO molecules for an indefinite period of time. The \( {}^{14}\mathrm{C} \) will be continuously transferred between CO and CO2 molecules as the forward and reverse reactions occur at the equilibrium state, making it present in both CO and CO2 molecules.

Step by step solution

01

Understand the concept of equilibrium

A chemical equilibrium is a state in a chemical reaction where the concentrations of the reactants and products do not change with time. This means that the reaction is happening at the same rate in both the forward and reverse directions, and the concentrations of all chemicals involved remain constant.
02

Analyze the role of radioactive isotopes in a chemical equilibrium

In this case, we have radioactive \( {}^{14}\mathrm{C} \) labeled in the CO molecule. At equilibrium, the CO molecules will react with H2O molecules to form H2 and CO2 molecules, and the reverse reaction will also occur. Since radioactive \( {}^{14}\mathrm{C} \) is part of the CO molecule initially, it will take part in the chemical reaction and will be found in newly formed CO2 molecules as well.
03

Determine the fate of \({}^{14}\mathrm{C}\) in the equilibrium

At equilibrium, the concentrations of all chemicals involved remain constant, but individual molecules are constantly exchanging between reactants and products. As the forward and reverse reactions proceed, radioactive \( {}^{14}\mathrm{C} \) will be transferred between CO and CO2 molecules due to the continuous reactions happening at the equilibrium state.
04

Conclusion

Radioactive \( {}^{14}\mathrm{C} \) will not be found only in CO molecules for an indefinite period of time. Instead, it will be continuously transferred between CO and CO2 molecules as the forward and reverse reactions occur at the equilibrium state.

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Most popular questions from this chapter

For the reaction: $$ 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g) $$ \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 M\), calculate the equilibrium \(\mathrm{O}_{3}\) concentration.

Le Châtelier's principle is stated (Section \(13.7)\) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

For the following reaction at a certain temperature $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ it is found that the equilibrium concentrations in a \(5.00\) -L rigid container are \(\left[\mathrm{H}_{2}\right]=0.0500 M,\left[\mathrm{~F}_{2}\right]=0.0100 M\), and \([\mathrm{HF}]=\) \(0.400 \mathrm{M}\). If \(0.200\) mole of \(\mathrm{F}_{2}\) is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ equilibrium constants ( \(K_{\mathrm{p}}\) values) as a function of temperature are \(300^{\circ} \mathrm{C}, \quad 4.34 \times 10^{-3}\) \(500^{\circ} \mathrm{C}, \quad 1.45 \times 10^{-5}\) \(600^{\circ} \mathrm{C}, \quad 2.25 \times 10^{-6}\) Is the reaction exothermic or endothermic?

The following equilibrium pressures were observed at a certain temperature for the reaction $$ \begin{array}{c} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\ P_{\mathrm{NH}_{3}}=3.1 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{~atm} \\ P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{~atm} \end{array} $$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=0.0167 \mathrm{~atm}\), and \(\underline{P_{\mathrm{H}}}=0.00761\) atm, does this represent a system at equilibrium?
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