Write the equilibrium expressions \(K_{c}\) for the following reactions. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Short Answer

Expert verified
Equilibrium expressions are: (a) \(K_c = \frac{[\mathrm{COCl}] [\mathrm{Cl}]}{[\mathrm{CO}] [\mathrm{Cl}_2]}\), (b) \(K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2] [\mathrm{Br}_2]}\), (c) \(K_c = \frac{[\mathrm{SO}_{2}]^2 [\mathrm{H}_{2} \mathrm{O}]^2}{[\mathrm{H}_{2} \mathrm{S}]^2 [\mathrm{O}_2]^3}\).

Step by step solution

01

- Understanding Equilibrium Constant (\(K_c\)) Expression

The equilibrium constant expression, denoted by the symbol \(K_c\), is calculated for a reaction at equilibrium by dividing the product of the concentrations of the products raised to their stoichiometric coefficients by the product of the concentrations of the reactants raised to their stoichiometric coefficients. The concentrations are measured in moles per liter (M). Gases are represented with (g) after their formula.
02

- Write the Equilibrium Expression for Reaction (a)

Start by writing the product concentrations raised to their stoichiometric coefficients over the reactant concentrations raised to their stoichiometric coefficients. For the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})\), the equilibrium expression is \[K_c = \frac{[\mathrm{COCl}] [\mathrm{Cl}]}{[\mathrm{CO}] [\mathrm{Cl}_2]}\]
03

- Write the Equilibrium Expression for Reaction (b)

For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})\), the equilibrium expression is \[K_c = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_2] [\mathrm{Br}_2]}\]Note that the concentration of \(\mathrm{HBr}\) is squared because it is produced with a stoichiometric coefficient of 2.
04

- Write the Equilibrium Expression for Reaction (c)

For the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), although it is given as an equation, we must still write it as an equilibrium expression. So, the equilibrium expression is \[K_c = \frac{[\mathrm{SO}_{2}]^2 [\mathrm{H}_{2} \mathrm{O}]^2}{[\mathrm{H}_{2} \mathrm{S}]^2 [\mathrm{O}_2]^3}\]For this reaction, square both \(\mathrm{SO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) because of their coefficients and raise \(\mathrm{O}_2\) to the power of 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Imagine a dance floor where dancers are constantly pairing up and separating at the same rate; the population of dancing couples remains consistent over time. This is a visual way of understanding chemical equilibrium. When a chemical reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction, the concentrations of reactants and products remain constant. It's important to clarify that this doesn't mean the reactions have stopped; they are still occuring, but their effects balance each other out.

At equilibrium, the system's properties, such as pressure, color, and concentration, won't show any net change despite the ongoing microscopic dance. This dynamic balance is crucial in chemical processes and product formation.
Reaction Quotient
How close is a chemical system to reaching equilibrium? That's where the reaction quotient, denoted by Q, steps in. This nifty tool helps us forecast the direction in which a reaction will proceed to achieve equilibrium. The reaction quotient is calculated just like the equilibrium constant, using the same concentration expression, but with the current concentrations instead of those at equilibrium.

The comparison of Q to the equilibrium constant, Kc, can tell us whether a reaction needs to shift to produce more reactants or products. If Q is less than Kc, the reaction moves forward to produce more products. If Q is greater, it will shift to produce more reactants.
Le Chatelier's Principle
Ever wondered what happens when external conditions are altered on a system at equilibrium? Henri Louis Le Chatelier had the same question and deduced a principle about it. Le Chatelier's Principle states that when a system at chemical equilibrium experiences a change in concentration, temperature, volume, or pressure, it will adjust to counteract the imposed change and restore a new equilibrium.

For instance, adding more reactant to an equilibrium system causes it to produce additional products. Similarly, increasing the temperature for an exothermic reaction will cause the system to produce more reactants. Understanding this concept allows chemists to optimize conditions for desired product yields in industrial reactions.
Stoichiometry
If chemical reactions were recipes, stoichiometry would be the section about the amount of each ingredient. Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It's all about the mole ratios derived from a balanced chemical equation, which tell us how much of each substance is needed or produced.

In essence, stoichiometry provides the mathematical framework behind reactions, ensuring that atoms are conserved and allowing us to predict the amounts of products formed. This is essential in various fields, from designing chemical reactors to determining the fuel efficiency of vehicles.

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Most popular questions from this chapter

Use Le Chatelier's principle to predict the effect that the change given in the first column of the following table has on the quantity in the second column for the following equilibrium system: $$ \begin{gathered} 5 \mathrm{CO}(\mathrm{g})+\mathrm{I}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{I}_{2}(\mathrm{~g})+5 \mathrm{CO}_{2}(\mathrm{~g}) \\\ \Delta H^{\circ}=-1175 \mathrm{~kJ} \end{gathered} $$

Write the reaction quotients \(Q_{c}\) for (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CuCl}_{2}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) (s) \(\rightarrow \mathrm{N}_{2} \mathrm{O}\) (g) \(+2 \mathrm{H}_{2} \mathrm{O}\) (g) (c) \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}\) (s) \(+\mathrm{CO}_{2}\) (g)

Consider the equilibrium \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). (a) If the partial pressure of \(\mathrm{CO}_{2}\) is increased, what happens to the partial pressure of \(\mathrm{H}_{2}\) ? (b) If the partial pressure of \(\mathrm{CO}\) is decreased, what happens to the partial pressure of \(\mathrm{CO}_{2}\) ? (c) If the concentration of \(\mathrm{CO}\) is increased, what happens to the concentration of \(\mathrm{H}_{2}\) ? (d) If the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is decreased, what happens to the equilibrium constant for the reaction?

Determine \(K_{c}\) from the value of \(K\) for cach of the following equilibria. (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.4\) at \(1000 \mathrm{~K}\) (b) \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}), K=9.4 \times 10^{-2}\) at \(24^{\circ} \mathrm{C}\)

State whether reactants or products will be favored by compression in each of the following equilibria. If no change occurs, explain why that is so. (a) \(2 \mathrm{O}_{3}(\mathrm{~g})=3 \mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(2 \mathrm{HD}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2}(\mathrm{~g})\) (e) \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{Cl}(\mathrm{g})\)

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