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Chapter 15: Problem 42

What is meant by "the position of an equilibrium"? Does the addition of a catalyst have any effects on the position of an equilibrium?

Short Answer

Expert verified
The position of an equilibrium refers to a state in a reversible reaction where the speeds of the forward and backward reactions are equal and the relative concentrations of reactants and products are constant. While a catalyst speeds up both forward and backward reactions, helping the system reach equilibrium faster, it does not change the position of the equilibrium.

Step by step solution

01

Understanding the position of an equilibrium

The position of an equilibrium can be defined as the state of a reversible reaction where the speeds of the forward and backward reactions have become equal. It's the relative concentrations of reactants and products in a reaction that have reached a balance, remaining constant over time. This does not mean that the amounts of reactant and product are equal, but that they are in a fixed ratio to each other. When this balance occurs, the reaction has reached its equilibrium state.
02

Explaining the role of a catalyst

A catalyst is a substance that speeds up a chemical reaction by providing an alternative reaction pathway with a lower activation energy. This allows more of the reactant particles to have enough energy to react and consequently the reaction rate increases.
03

Impact of a catalyst on the position of an equilibrium

While a catalyst speeds up the reactions, it does so for both the forward and backward reactions equally. This means that a catalyst helps a system reach equilibrium more quickly, but it does not change the position of the equilibrium. The concentrations of reactants and products at equilibrium will be the same whether a catalyst is present or not.

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Most popular questions from this chapter

When a gas was heated under atmospheric conditions, its color was found to deepen. Heating above \(150^{\circ} \mathrm{C}\) caused the color to fade, and at \(550^{\circ} \mathrm{C}\) the color was barely detectable. However, at \(550^{\circ} \mathrm{C}\), the color was partially restored by increasing the pressure of the system. Which of these best fits this description? Justify your choice. (a) A mixture of hydrogen and bromine, (b) pure bromine, (c) a mixture of nitrogen dioxide and dinitrogen tetroxide. (Hint: Bromine has a reddish color and nitrogen dioxide is a brown gas. The other gases are colorless.)

The vapor pressure of mercury is \(0.0020 \mathrm{mmHg}\) at \(26^{\circ} \mathrm{C}\). (a) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the process \(\mathrm{Hg}(l) \rightleftharpoons \mathrm{Hg}(g) .\) (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring \(6.1 \mathrm{~m}\) long, \(5.3 \mathrm{~m}\) wide, and \(3.1 \mathrm{~m}\) high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor in \(\mathrm{mg} / \mathrm{m}^{3}\). Does this concentration exceed the safety limit of \(0.050 \mathrm{mg} / \mathrm{m}^{3} ?\) (Ignore the volume of furniture and other objects in the laboratory.)

Photosynthesis can be represented by $$ \begin{array}{r} 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \\ \Delta H^{\circ}=2801 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Explain how the equilibrium would be affected by the following changes: (a) partial pressure of \(\mathrm{CO}_{2}\) is increased, (b) \(\mathrm{O}_{2}\) is removed from the mixture, (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (sucrose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f) temperature is decreased, (g) more sunlight falls on the plants.

At \(25^{\circ} \mathrm{C}\), the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.15 \mathrm{~atm}\) and \(0.20 \mathrm{~atm}\), respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

The dissociation of molecular iodine into iodine atoms is represented as $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ At \(1000 \mathrm{~K}\), the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(I_{2}\) in a 2.30 - \(L\) flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?
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