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Learning Goal: To use the equilibrium constant and Le Châtelier's principle to determine how a reaction will respond to external factors. A reaction is at equilibrium when the concentrations of the reactants and products no longer change over time. This does not mean the reaction is over; rather, two competing reactions continue to occur simultaneously at equal rates. The two competing reactions are the forward reaction (reactants \( \rightarrow \) products) and the reverse reaction (products \( \rightarrow \) reactants). If a reaction at equilibrium is subjected to a stress, the concentrations of reactants and products adjust to reestablish equilibrium. This is called Le Châtelier's principle. A stress might be a change in the concentration of reactants or products, a change in the volume of the reaction container, a change in temperature, or the addition of a catalyst.

Ask by Bowen Erickson. in the United States
Mar 03,2025

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A reaction is at equilibrium when the forward and reverse reactions happen at the same rate, keeping the amounts of reactants and products steady. If something changes, like adding more reactants or changing the temperature, the system adjusts to balance things out again. This is called Le Châtelier's principle.

Solution

This passage explains some key ideas about chemical equilibrium and how systems respond to external changes. Here are the main points in simpler terms: 1. A reaction is at equilibrium when the rates of the forward and reverse reactions are equal. This means that even though the reactions are still happening, the concentrations of all the reactants and products stay constant over time. 2. Equilibrium does not mean that the reaction has “stopped” or that the reactants have been completely used up. Instead, it means that the system has reached a balance where the forward reaction (making products from reactants) occurs at the same rate as the reverse reaction (making reactants from products). 3. Le Châtelier's principle tells us that if a change (or “stress”) is applied to a system at equilibrium, the system will adjust in order to counteract that change and establish a new equilibrium. Common stresses include:   • Changing the concentrations of reactants or products (e.g., adding more of a reactant).   • Changing the volume of the reaction container (which affects pressure, especially in reactions involving gases).   • Changing the temperature.   • Adding a catalyst (although it's important to note that a catalyst speeds up both the forward and reverse reactions equally, so it doesn’t change the overall equilibrium composition). Understanding these concepts helps predict how a system at equilibrium will shift when any of these factors change.

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The Deep Dive

Did you know that the concept of equilibrium dates back over 150 years to the work of chemists like Cato Guldberg and Peter Waage? They formulated the law of mass action in 1864, which laid the groundwork for our modern understanding of chemical equilibrium. This principle emphasizes that at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, allowing scientists to predict the behavior of reactions in countless chemical processes in industries today! Now, let’s tackle a common misconception: many students think that adding a catalyst will shift the equilibrium position. However, that’s not the case! Catalysts speed up the rate at which equilibrium is reached but do not change the position of equilibrium itself. Remember, only changes in concentration, pressure, or temperature will influence where that equilibrium lies. So, when in doubt, think about how those stresses will push your reaction to the right or left!

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1. Calculate the number of molecules in 0.75 mol of \( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \). 2. Calculate the number of Oxygen atoms in a 0.350 mol of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \). 3. Sodium carbonate has \( 2.41 \times 10^{24} \) molecules. Calculate the mass of sodium carbonate. 4. Determine the empirical formula of a compound composed of \( 62 \% \mathrm{C}, 5.20 \% \mathrm{H}, 12.1 \% \) N , and \( 20.6 \% \mathrm{O} \). 5. Combustion analysis of 3.58 mg ethyl butyrate produces, \( 7.12 \mathrm{mg} \mathrm{CO}_{2} \), and 3.58 mg \( \mathrm{H}_{2} \mathrm{O} \). Calculate the mass of oxygen in ethyl butyrate. 6. A mass of 5.915 g of a compound containing on \( \mathrm{C}, \mathrm{H} \) and O was burned in excess oxygen. The combustion reaction resulted in \( 11.942 \mathrm{~g} \mathrm{CO}_{2} \) and 0.749 g of \( \mathrm{H}_{2} \mathrm{O} \). Determine the empirical formula of the compound. 7. A compound \( \mathrm{C}_{3} \mathrm{H}_{8} \) weighing 4.54 g undergoes combustion according to the reaction: \( \mathrm{C}_{3} \mathrm{H}_{8}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \) a. Balance the given equation. b. How much in grams is \( \mathrm{CO}_{2} \) produced. 8. Balance the following equations: a. \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \rightarrow \mathrm{PbO}+\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \) b. \( \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \rightarrow \quad \mathrm{N}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad+\quad \mathrm{O}_{2}(\mathrm{~g}) \) 9. The following reaction was conducted in the lab, here is the balanced equation: \[ \mathrm{K}_{2} \mathrm{CrO}_{4}(\mathrm{aq})+\quad 2 \mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow \quad \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{~s})+\quad 2 \mathrm{KNO}_{3}(\mathrm{aq}) \] A mass of \( 1.345 \mathrm{~g} \mathrm{~K}_{2} \mathrm{CrO}_{4} \) was reacted with \( 2.536 \mathrm{~g} \mathrm{AgNO}_{3} \). a. Find the limiting reactant.

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1. Calculate the number of molecules in 0.75 mol of \( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \). 2. Calculate the number of Oxygen atoms in a 0.350 mol of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \). 3. Sodium carbonate has \( 2.41 \times 10^{24} \) molecules. Calculate the mass of sodium carbonate. 4. Determine the empirical formula of a compound composed of \( 62 \% \mathrm{C}, 5.20 \% \mathrm{H}, 12.1 \% \) N , and \( 20.6 \% \mathrm{O} \). 5. Combustion analysis of 3.58 mg ethyl butyrate produces, \( 7.12 \mathrm{mg} \mathrm{CO}_{2} \), and 3.58 mg \( \mathrm{H}_{2} \mathrm{O} \). Calculate the mass of oxygen in ethyl butyrate. 6. A mass of 5.915 g of a compound containing on \( \mathrm{C}, \mathrm{H} \) and O was burned in excess oxygen. The combustion reaction resulted in \( 11.942 \mathrm{~g} \mathrm{CO}_{2} \) and 0.749 g of \( \mathrm{H}_{2} \mathrm{O} \). Determine the empirical formula of the compound. 7. A compound \( \mathrm{C}_{3} \mathrm{H}_{8} \) weighing 4.54 g undergoes combustion according to the reaction: \( \mathrm{C}_{3} \mathrm{H}_{8}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \) a. Balance the given equation. b. How much in grams is \( \mathrm{CO}_{2} \) produced. 8. Balance the following equations: a. \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \rightarrow \mathrm{PbO}+\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \) b. \( \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s}) \rightarrow \quad \mathrm{N}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad+\quad \mathrm{O}_{2}(\mathrm{~g}) \) 9. The following reaction was conducted in the lab, here is the balanced equation: \[ \mathrm{K}_{2} \mathrm{CrO}_{4}(\mathrm{aq})+\quad 2 \mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow \quad \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{~s})+\quad 2 \mathrm{KNO}_{3}(\mathrm{aq}) \] A mass of \( 1.345 \mathrm{~g} \mathrm{~K}_{2} \mathrm{CrO}_{4} \) was reacted with \( 2.536 \mathrm{~g} \mathrm{AgNO}_{3} \). a. Find the limiting reactant.
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