Chapter 15, Problem 15.155QP

For which of the following reactions is K_c equal to K_p? For which can we not write a K_p expression?

$\begin{array}{l}\left(\text{a}\right){\text{4NH}}_{\text{3}}\left(g\right)+{\text{5O}}_{\text{2}}\left(g\right)\rightleftarrows \text{4NO}\left(g\right)+{\text{6H}}_{\text{2}}\text{O}\left(g\right)\\ \left(\text{b}\right){\text{CaCO}}_{\text{3}}\left(s\right)\rightleftarrows \text{CaO}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)\\ \left(\text{c}\right)\text{Zn}\left(s\right)+{\text{2H}}^{\text{+}}\left(aq\right)\rightleftarrows {\text{Zn}}^{\text{2+}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)\\ \left(\text{d}\right){\text{PCl}}_{\text{3}}\left(g\right)+{\text{3NH}}_{\text{3}}\left(g\right)\rightleftarrows \text{3HCl}\left(g\right)+\text{P}{\left({\text{NH}}_{\text{2}}\right)}_3\left(g\right)\\ \left(\text{e}\right){\text{NH}}_{\text{3}}\left(g\right)+\text{HCl}\left(g\right)\rightleftarrows {\text{NH}}_{\text{4}}\text{Cl}\left(s\right)\\ \left(\text{f}\right){\text{NaHCO}}_{\text{3}}\left(s\right)+{\text{H}}^{\text{+}}\left(aq\right)\rightleftarrows {\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{Na}}^{\text{+}}\left(aq\right)\\ \left(g\right){\text{H}}_{\text{2}}\left(g\right)+{\text{F}}_{\text{2}}\left(g\right)\rightleftarrows \text{2HF}\left(g\right)\\ \left(h\right)\text{C}\left(\text{graphite}\right)+{\text{CO}}_{\text{2}}\left(g\right)\rightleftarrows \text{2CO}\left(g\right)\end{array}$

Interpretation Introduction

Interpretation:

The enthalpy values $\text{(}\pm \text{Δn)}$ should be calculated identified given the different equilibrium reactions, with supporting of equilibrium equation.

Concept Introduction:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Equilibrium pressure: The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.

Homogeneous equilibrium: A homogeneous equilibrium involved has a everything present in the same phase and same conditions, for example reactions where everything is a gas, or everything is present in the same solution.

Heterogeneous equilibrium: This equilibrium reaction does not depend on the amounts of pure solid and liquid present, in other words heterogeneous equilibrium, substances are in different phases.

Forward Reaction: This type of reaction has involved irreversible, if obtained product cannot be converted back in to respective reactants under the same conditions. Backward Reaction: This type of reaction process involved a reversible, if the products can be converted into a back to reactants.

Answer to Problem 15.155QP

The fallowing equilibrium (Kp and Kc) process is the given the solid, liquid phase reactions are shown below.

$\begin{array}{l}\text{a)}\text{.}{\text{4NH}}_{\text{3}}\text{(g)}\text{+}{\text{5O}}_{\text{2}}\text{(g)}\rightleftharpoons 4\text{NO(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(g)-------(}\Delta \text{n}\ne \text{0)}\\ b).CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)-------\text{(}\Delta \text{n}\ne \text{0)}\\ c).Zn(s)+{\text{2H}}^{\text{+}}\text{(aq})\rightleftharpoons Z{n}^{2+}(aq)+{\text{H}}_{\text{2}}\text{(g})-----\text{Some}\text{species}\text{are}\text{aqueous}\\ d).PC{l}_3(g)+3N{H}_3(g)\rightleftharpoons 3HCl(g)+{\text{P(NH}}_{\text{2}}{\text{)}}_{\text{3}}(g)---\text{(}\Delta \text{n=0)}\\ e).N{H}_3(g)+HCl(g)\rightleftharpoons N{H}_{4}Cl(s)----------\text{(}\Delta \text{n}\ne \text{0)}\\ f).NaHC{O}_3(s)+H^{+}(aq)\rightleftharpoons H_{2}O(l)+C{O}_2(g)+N{a}^{+}(aq)----\text{Some}\text{species}\text{are}\text{aqueous}\\ g).H_2(g)+{\text{F}}_{\text{2}}\text{(g})\rightleftharpoons 2HF(g)---------\text{(}\Delta \text{n=0)}\\ h).C(graphite)+CO(g)\rightleftharpoons 2CO(g)------\text{(}\Delta \text{n}\ne \text{0)}\\ \end{array}$ To Identify: The given equilibrium reactions (a-d) are $\text{(}\pm \text{Δn)}$ values should be calculated and analyzed

The number of products should be calculated with given below equilibrium equation.

Reaction (a)

$\begin{array}{l}\text{a)}\text{.}{\text{4NH}}_{\text{3}}\text{(g)}\text{+}{\text{5O}}_{\text{2}}\text{(g)}\rightleftharpoons 4\text{NO(g)}\text{+}{\text{6H}}_{\text{2}}\text{O(g)}\\ \text{Kc}\text{=}\frac{{[NO]}^4{[H_{2}O]}^6}{{[N{H}_3]}^4{[O_2]}^5}\\ \text{Kp}\text{=}\frac{\text{Product}}{\text{Reactant}}=\frac{{(P_{NO})}^4{(P_{H_{2}O})}^6}{{(P_{N{H}_3})}^4{(P_{O_2})}^5}\\ \text{Kp}\text{= }[Weconsidergasphasemolecule\text{only}]\\ \text{Consider}\text{the following}\text{equation}\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ \text{Kp}\text{=}\text{Kc}{\text{[0}}^{\text{.08206}}\\ \text{Kp}\text{=}\text{Kc}{(}^{\text{0}}=\text{Kc}\cdot \text{1=Kc}\\ \text{Temprature}\text{ 273K }\\ \text{Here}\text{Δn}\text{=}10\text{-9}\text{=}+1\text{(Product - Reactant) }\\ =\Delta n=+1\\ \text{Kp}\text{=}\text{Kc}\text{(}\Delta \text{n}\ne \text{0)}\\ \end{array}$

Reaction (b):

$\begin{array}{l}b).CaC{O}_3(s)\rightleftharpoons CaO(s)+C{O}_2(g)\\ \text{Δn}\text{=}1\text{-0=}+1\text{(Product - Reactant) }\\ =\Delta n=+1\\ \text{Kp}\text{=}\text{Kc}\text{(}\Delta \text{n}\ne \text{0)}\\ \text{Kp}\text{=}\text{Kc}\text{(}\Delta \text{n}\ne \text{0)}\text{Kp}\text{= }[Weconsidergasphasemolecule\text{only}]\end{array}$

Reaction (c)

$\begin{array}{l}c).Zn(s)+{\text{2H}}^{\text{+}}\text{(aq})\rightleftharpoons Z{n}^{2+}(aq)+{\text{H}}_{\text{2}}\text{(g})\\ \text{This }\text{reactions should}\text{not}\text{be}\text{calculated (Kp)}\text{values}\\ \text{Because}\text{some}\text{species}\text{are}\text{aqueous}\end{array}$

Reaction (d)

$\begin{array}{l}d).PC{l}_3(g)+3N{H}_3(g)\rightleftharpoons 3HCl(g)+{\text{P(NH}}_{\text{2}}{\text{)}}_{\text{3}}(g)\\ \text{Kp}\text{=}\frac{(P_{PC{l}_3}){(P_{N{H}_3})}^3}{{(P_{HCl})}^3(P_{{\text{P(NH}}_{\text{2}}{\text{)}}_{\text{3}}})}\\ \text{Consider}\text{the following}\text{equation}\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ \text{Kp}\text{=}\text{Kc}{\text{[0}}^{\text{.08206}}\\ \text{Kp}\text{=}\text{Kc}{(}^{\text{0}}=\text{Kc}\cdot \text{1=Kc}\\ \text{Here}\text{Δn}\text{=}4\text{-0}\text{=}0\text{(Product - Reactant) }\\ =\Delta n=0\\ \text{Kp}\text{=}\text{Kc}\text{(}\Delta \text{n=0)}\end{array}$

Explanation of Solution

Analyzing equilibrium reaction: Given the different reactants molecule produced a two different product molecules, which is one gas another one liquid the balanced equation are placed above. Hence the concentration of solids and pure liquids does not appear in molar constant (Kc), so given multiple equilibrium equations are (a and d) homogenies reactions meanwhile reaction (b and c) heterogonous.

Further we derived number of reactant and product values are presented above, with respective reactions.

To Identify: The given equilibrium reactions (e-h) are $\text{(}\pm \text{Δn)}$ values should be calculated and analyzed

The number of products should be calculated with given below equilibrium equation.

Reaction (e)

$\begin{array}{l}e).N{H}_3(g)+HCl(g)\rightleftharpoons N{H}_{4}Cl(s)----------\text{(}\Delta \text{n}\ne \text{0)}\\ \text{Kp}\text{= }[Weconsidergasphasemolecule\text{only}]\\ \text{Consider}\text{the following}\text{equation}\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ \text{Kp}\text{=}\text{Kc}{\text{[0}}^{\text{.08206}}\\ \text{Kp}\text{=}\text{Kc}{(}^{\text{0}}=\text{Kc}\cdot \text{1=Kc}\\ \text{Here}\text{Δn}\text{=}0\text{-2}\text{=}-2\text{(Product - Reactant) }\\ =\Delta n=-2\\ \text{Kp}\text{=}\text{Kc}\text{(}\Delta \text{n}\ne \text{0)}\end{array}$

Reaction (f)

$\begin{array}{l}f).NaHC{O}_3(s)+H^{+}(aq)\rightleftharpoons H_{2}O(l)+C{O}_2(g)+N{a}^{+}(aq)\\ \text{This }\text{reactions should}\text{not}\text{be}\text{calculated (Kp)}\text{values}\\ \text{Because}\text{some}\text{species}\text{are}\text{aqueous}\end{array}$

Reaction (g)

$\begin{array}{l}g).H_2(g)+{\text{F}}_{\text{2}}\text{(g})\rightleftharpoons 2HF(g)\\ \text{Kp}\text{=}\frac{(P_{PC{l}_3}){(P_{N{H}_3})}^3}{{(P_{HCl})}^3(P_{{\text{P(NH}}_{\text{2}}{\text{)}}_{\text{3}}})}\\ \text{Consider}\text{the following}\text{equation}\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ \text{Kp}\text{=}\text{Kc}{\text{[0}}^{\text{.08206}}\\ \text{Kp}\text{=}\text{Kc}{(}^{\text{0}}=\text{Kc}\cdot \text{1=Kc}\\ \text{Here}\text{Δn}\text{=}2\text{-2}\text{=}0\text{(Product - Reactant) }\\ =\Delta n=0\\ \text{Kp}\text{=}\text{Kc}\text{(}\Delta \text{n=0)}\end{array}$

Reaction (h)

$\begin{array}{l}h).C(graphite)+CO(g)\rightleftharpoons 2CO(g)------\text{(}\Delta \text{n}\ne \text{0)}\\ \text{Kp}\text{= }[Weconsidergasphasemolecule\text{only}]\\ \text{Consider}\text{the following}\text{equation}\\ Kc=\frac{\text{Kp}}{{\text{(RT)}}^{\text{Δn}}}(or)Kp={\text{(RT)}}^{\text{Δn}}\\ \text{Kp}\text{=}\text{Kc}{\text{[0}}^{\text{.08206}}\\ \text{Kp}\text{=}\text{Kc}{(}^{\text{0}}=\text{Kc}\cdot \text{1=Kc}\\ \text{Here}\text{Δn}\text{=}2\text{-1}\text{=}+1\text{(Product - Reactant) }\\ =\Delta n=+1\\ \text{Kp}\text{=}\text{Kc}\text{(}\Delta \text{n}\ne \text{0)}\end{array}$

Explanation:

Analyzing equilibrium reaction: Given the different reactants molecule produced a two different product molecules, which is one gas another one liquid the balanced equation are placed above. Hence the concentration of solids and pure liquids does not appear in molar constant (Kc), so given multiple equilibrium equations are (g) homogenies reactions meanwhile reaction (e, f, and h) heterogonous.

Further we derived number of reactant and product values are presented above, with respective reactions.

Conclusion:

The $(\pm \Delta n)$ value is solved given the heterogeneous reaction with supporting of (Kp and Kc) equilibrium equations.