/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 The reaction $$ 2 \mathrm{PH... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 63

The reaction $$ 2 \mathrm{PH}_{3}(g)+\mathrm{As}_{2}(g) \rightleftharpoons 2 \mathrm{AsH}_{3}(g)+\mathrm{P}_{2}(g) $$ has \(K_{\mathrm{p}}=2.9 \times 10^{-5}\) at \(873 \mathrm{~K}\). At the same temperature, what is \(K_{\mathrm{p}}\) for each of the following reactions? (a) \(2 \mathrm{AsH}_{3}(g)+\mathrm{P}_{2}(g) \rightleftharpoons 2 \mathrm{PH}_{3}(g)+\mathrm{As}_{2}(g)\) (b) \(6 \mathrm{PH}_{3}(g)+3 \mathrm{As}_{2}(g) \rightleftharpoons 3 \mathrm{P}_{2}(g)+6 \mathrm{AsH}_{3}(g)\) (c) \(2 \mathrm{P}_{2}(g)+4 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}_{2}(g)+4 \mathrm{PH}_{3}(g)\)

Short Answer

Expert verified
(a) \(K_{p(a)} = 3.45 \times 10^4\), (b) \(K_{p(b)} = 2.44 \times 10^{-14}\), (c) \(K_{p(c)} = 1.19 \times 10^9\).

Step by step solution

01

Consider Reaction (a) - Reverse Reaction

The first reaction given is the reverse of the original reaction. The equilibrium constant for a reverse reaction is the inverse of the forward reaction's equilibrium constant. Thus, \(K_{p(a)} = \frac{1}{K_p} = \frac{1}{2.9 \times 10^{-5}}\).
02

Calculate Kp for Reaction (a)

Calculate the equilibrium constant for reaction (a) by taking the inverse of the given \(K_p\):\[K_{p(a)} = \frac{1}{2.9 \times 10^{-5}} \approx 3.45 \times 10^4\].
03

Consider Reaction (b) - Multiples of Reaction

Reaction (b) is the original reaction multiplied by a factor of 3. For reactions scaled by a factor \(n\), the equilibrium constant becomes \((K_p)^n\). Here, \(n=3\), so \(K_{p(b)} = (K_p)^3\).
04

Calculate Kp for Reaction (b)

Raise the given \(K_p\) to the power of 3:\[K_{p(b)} = (2.9 \times 10^{-5})^3 = 2.44 \times 10^{-14}\].
05

Consider Reaction (c) - Reverse and Scale Reaction

Reaction (c) is the reverse of the original reaction, scaled by a factor of 2. Reverse first, which gives the inverse \(K_p\), and then raise it to the power 2. So, \(K_{p(c)} = \left(\frac{1}{K_p}\right)^2\).
06

Calculate Kp for Reaction (c)

Take the inverse of \(K_p\) and square it:\[K_{p(c)} = \left(\frac{1}{2.9 \times 10^{-5}}\right)^2 = (3.45 \times 10^4)^2 = 1.19 \times 10^9\].

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Key Concepts

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

Reversible Reactions
In the world of chemistry, many reactions can proceed in both forward and backward directions. These are called *reversible reactions*.
For a reaction that can go both ways, you will often see a double-headed arrow (\(\rightleftharpoons\)) in the equation.
Reversible reactions don't just stop once the products are formed. The products can convert back into reactants too.
This back and forth keeps happening until a balance is reached. This balance is known as equilibrium. It's when the rate of the forward and backward reactions are equal, and the concentrations of the reactants and products remain constant.The concept of reversibility is key because it leads directly to the understanding of the equilibrium constant, \(K_p\). In simple terms, **for a reversible reaction:**
  • The equilibrium constant for the reverse reaction is the inverse of that for the forward reaction.
  • If the reaction is scaled by a factor, the equilibrium constant must also be adjusted accordingly.
Reaction Stoichiometry
Stoichiometry is the heart of understanding how reactions work in chemistry.
It involves studying the amounts of reactants and products involved in a chemical reaction.When it comes to reactions, coefficients in the balanced equation tell you how much of each substance is consumed or produced.
In stoichiometry, these coefficients help you calculate things like mass, moles, and the equilibrium constant.
For example, if you multiply the whole reaction by 2, you need to understand how that affects the equilibrium constant:\( (K_p)^n \) where \( n \) is the factor by which the reaction is multiplied.**Key Points of Reaction Stoichiometry:**
  • Chemical equations must be balanced to accurately reflect the stoichiometry of a reaction.
  • The equilibrium constant changes if the entire reaction equation is scaled by a factor. Raise \(K_p\) to the power of that factor to adjust for the change.
Thermodynamics
Thermodynamics is the study of energy changes in reactions and physical changes.
It enables us to understand whether a reaction is spontaneous or requires energy input. In the context of equilibria, thermodynamics helps in understanding how temperature, pressure, and concentration affect the equilibrium state. One of its applications is predicting the behavior of reactions at different temperatures. It also tells us how equilibrium constants are derived based on the energy changes in reactions.
This is particularly informative as *equilibrium constants* are temperature dependent. **Thermodynamics and Equilibrium:**
  • Equilibrium constants change with temperature because they are tied to the enthalpy and entropy of the system.
  • Knowledge of thermodynamics can help predict the direction of a shift in equilibrium when external conditions like temperature change.
Understanding these thermodynamic principles helps deeply in conjunction with studying equilibrium in reversible reactions.

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Most popular questions from this chapter

The affinity of hemoglobin (Hb) for CO is greater than its affinity for \(\mathrm{O}_{2}\). Use Le Ch芒telier's principle to predict how CO affects the equilibrium \(\mathrm{Hb}+\mathrm{O}_{2} \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right) .\) Suggest a possible reason for the toxicity of CO.

At \(298 \mathrm{~K}, K_{c}\) is \(2.2 \times 10^{5}\) for the reaction \(\mathrm{F}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{O}_{2} \mathrm{~F}(g) .\) What is the value of \(K_{\mathrm{p}}\) at this temperature?

The first step in the industrial synthesis of hydrogen is the reaction of steam and methane to give synthesis gas, a mixture of carbon monoxide and hydrogen: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CH}_{4}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \quad K_{c}=4.7\) at \(1400 \mathrm{~K}\) A mixture of reactants and products at \(1400 \mathrm{~K}\) contains \(0.035\) \(\mathrm{M} \mathrm{H}_{2} \mathrm{O}, 0.050 \mathrm{M} \mathrm{CH}_{4}, 0.15 \mathrm{M} \mathrm{CO}\), and \(0.20 \mathrm{M} \mathrm{H}_{2}\). In which di- rection does the reaction proceed to reach equilibrium?

Will the concentration of \(\mathrm{NO}_{2}\) increase, decrease, or remain the same when the equilibrium $$ \mathrm{NO}_{2} \mathrm{Cl}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NOCl}(g)+\mathrm{NO}_{2}(g) $$ is disturbed by the following changes? (a) Adding \(\mathrm{NOCl}\) (b) Adding \(\mathrm{NO}\) (c) Removing \(\mathrm{NO}\) (d) Adding \(\mathrm{NO}_{2} \mathrm{Cl} ;\) also account for the change using the reaction quotient \(Q_{c}\)

The value of \(K_{c}\) for the reaction of acetic acid with ethanol is \(3.4\) at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{soln})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\text { soln }) \rightleftharpoons $$ Aceticacid $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}(s o l n)+\mathrm{H}_{2} \mathrm{O}(\text { soln }) \quad K_{c}=3.4 \\ &\text { Ethyl acetate } \end{aligned} $$ (a) How many moles of ethyl acetate are present in an equilibrium mixture that contains \(4.0 \mathrm{~mol}\) of acetic acid, \(6.0 \mathrm{~mol}\) of ethanol, and \(12.0 \mathrm{~mol}\) of water at \(25^{\circ} \mathrm{C}\) ? (b) Calculate the number of moles of all reactants and products in an equilibrium mixture prepared by mixing \(1.00 \mathrm{~mol}\) of acetic acid and \(10.00 \mathrm{~mol}\) of ethanol.
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