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Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) c. \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) d. \(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
a. \(K_{\mathrm{p}}=\frac{1}{[\mathrm{O}_{2}]^{\frac{3}{2}}}\) b. \(K_{\mathrm{p}}=[\mathrm{CO}_{2}]\) c. \(K_{\mathrm{p}}=\frac{[\mathrm{CO}][\mathrm{H}_{2}]}{[\mathrm{H}_{2}\mathrm{O}]}\) d. \(K_{\mathrm{p}}=\frac{[\mathrm{O}_{2}]^3}{[\mathrm{H}_{2}\mathrm{O}]^2}\)

Step by step solution

01

Identify the gaseous species

In this reaction, we have O鈧(g) as the only gaseous species involved.
02

Expression for Kp

To write the expression for Kp for this reaction, we have: \[K_{\mathrm{p}}=\frac{1}{[\mathrm{O}_{2}]^{\frac{3}{2}}}\] b. $\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)$
03

Identify the gaseous species

In this reaction, we have CO鈧(g) as the only gaseous species involved.
04

Expression for Kp

To write the expression for Kp for this reaction, we have: \[K_{\mathrm{p}}=[\mathrm{CO}_{2}]\] c. $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)$
05

Identify the gaseous species

In this reaction, we have H鈧侽(g), CO(g), and H鈧(g) as the gaseous species involved.
06

Expression for Kp

To write the expression for Kp for this reaction, we have: \[K_{\mathrm{p}}=\frac{[\mathrm{CO}][\mathrm{H}_{2}]}{[\mathrm{H}_{2}\mathrm{O}]}\] d. $4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)$
07

Identify the gaseous species

In this reaction, we have H鈧侽(g) and O鈧(g) as the gaseous species involved.
08

Expression for Kp

To write the expression for Kp for this reaction, we have: \[K_{\mathrm{p}}=\frac{[\mathrm{O}_{2}]^3}{[\mathrm{H}_{2}\mathrm{O}]^2}\]

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Key Concepts

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

Kp expressions
The equilibrium constant in terms of partial pressures, denoted as \(K_{\mathrm{p}}\), is a way of expressing the equilibrium condition of a gaseous reaction. It takes into account only the gaseous species involved in the reaction, ignoring solids and pure liquids as they do not have a significant impact on the equilibrium constant. To write a \(K_{\mathrm{p}}\) expression, you use the partial pressures of the gaseous reactants and products.
For a general reaction like \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\), the \(K_{\mathrm{p}}\) expression is:\[K_{\mathrm{p}} = \frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b}\]where \(P_A\), \(P_B\), \(P_C\), and \(P_D\) denote the partial pressures of the respective gases, and \(a\), \(b\), \(c\), \(d\) are their stoichiometric coefficients in the balanced equation.
In practice, when writing a \(K_{\mathrm{p}}\) expression:- Include only gaseous species.- Apply coefficients as exponents in the expression.
gaseous species
Gaseous species in a chemical reaction are the components that exist in the gas phase under the conditions of the reaction. These are crucial when determining expressions like \(K_{\mathrm{p}}\), as only these species are included in the calculations. In contrast, solids and pure liquids do not change significantly in concentration or pressure, so they are typically omitted from \(K_{\mathrm{p}}\) expressions.
To identify gaseous species:- Look at the state symbol next to each chemical formula in the equation, "(g)" usually indicates a gas.- Consider the equilibrium conditions to ensure these compounds are indeed gases at the prevailing temperature and pressure.Recognizing these species helps in constructing meaningful equilibrium expressions, crucial for predicting how changes in conditions will affect the system.
In the given exercises:- Example "a" includes \(\mathrm{O}_2(g)\).- Example "b" includes \(\mathrm{CO}_2(g)\).- Example "c" involves \(\mathrm{H}_2\mathrm{O}(g)\), \(\mathrm{CO}(g)\), and \(\mathrm{H}_2(g)\).- Example "d" has \(\mathrm{H}_2\mathrm{O}(g)\) and \(\mathrm{O}_2(g)\).The presence of these gaseous species dictates which terms appear in the equations for \(K_{\mathrm{p}}\).
chemical equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a closed system. At this point, the concentrations of reactants and products remain constant over time, not because the reactions stop occurring, but because they occur at an equal rate in both directions. This state of balance allows us to define equilibrium constants like \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}}\) depending on whether the concentrations or partial pressures are used.
Understanding equilibrium is vital since it explains how and why reactions reach a state where reactants and products coexist in a stable ratio.
Some characteristics of equilibrium include:- Dynamic nature: reactions continue to occur.- Reversible process: both forward and reverse reactions are possible.- Constant macroscopic properties: concentrations, color, pressure, and other observable factors don鈥檛 change at equilibrium.It is important in practical scenarios, where modifying conditions like temperature or pressure can shift the equilibrium position as described by Le Chatelier's principle.
reaction quotient
The reaction quotient, represented as \(Q\), is used to determine the direction a reaction will proceed to reach equilibrium. Its form resembles that of the equilibrium constant \(K_{\mathrm{p}}\) or \(K_{\mathrm{c}}\), but unlike \(K\) values, \(Q\) can be calculated at any point during the reaction, not just at equilibrium.
To find \(Q\):- Use the same expression as used for \(K_{\mathrm{p}}\) or \(K_{\mathrm{c}}\), but instead of equilibrium concentrations or pressures, insert the current concentrations or pressures of the reactants and products.Comparing \(Q\) with \(K\) helps in predicting the direction of the shift:
  • If \(Q < K\), the reaction will proceed forward, forming more products to reach equilibrium.
  • If \(Q = K\), the system is at equilibrium, and no net change will occur.
  • If \(Q > K\), the reaction will proceed in reverse, forming more reactants to reach equilibrium.
This tool offers valuable insight, especially when adjustments are made to the reaction conditions, guiding how the system can be manipulated to favor the formation of desired products.

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Most popular questions from this chapter

Calculate a value for the equilibrium constant for the reaction $$\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ given $$\mathrm{NO}_{2}(g) \stackrel{h \nu}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) \quad K=6.8 \times 10^{-49}$$ $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad K=5.8 \times 10^{-34}$$ (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$ Calculate the concentrations of all species at equilibrium for each of the following cases. a. 1.0 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and 2.0 \(\mathrm{g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a 1.0 -L flask. b. 1.0 mole of pure HOCl is placed in a 2.0 \(\mathrm{L}\) flask.

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) . \mathrm{A}\) friend asks the following: 鈥淚 know we have been told that if a mixture of A, B, C, and D is at equilibrium and more of A is added, more C and D will form. But how can more C and D form if we do not add more B?鈥 What do you tell your friend?

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.

An 8.00 -g sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ At equilibrium the total pressure and the density of the gaseous mixture were 1.80 \(\mathrm{atm}\) and 1.60 \(\mathrm{g} / \mathrm{L}\) , respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

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