Chapter 15.3, Problem 15.4WE

The following reactions have the indicated equilibrium constants at 100°C.

(1) 2NOBr(g) ⇄ 2NO(g) + Br₂(g) K_c = 0.014

(2) Br₂(g) + Cl₂(g) ⇄ 2BrCl(g) K_c = 7.2

Determine the value of K_c for the following reactions at 100°C.

(a) 2NO(g) + Br₂(g) ⇄ 2NOBr(g)

(b) 4NOBr(g) ⇄ 4NO(g) + 2Br₂(g)

(c) $\text{NOBr}\left(g\right)\rightleftarrows \text{NO}\left(g\right)+\frac{1}{2}{\text{Br}}_{\text{2}}(g)$

(d) 2NOBr(g) + Cl₂(g) ⇄ 2NO(g) + 2BiCl(g)

(e) $\text{NO}\left(g\right)\text{ + BrCl}\left(g\right)\rightleftarrows \text{NOBr}\left(g\right)+\frac{1}{2}{\text{Cl}}_{\text{2}}(g)$

Interpretation Introduction

Interpretation:

To determine the equilibrium constant (Kc) values should be calculate given the different equilibrium reactions at 100⁰C.

Concept Introduction:

Chemical equilibrium: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved when the concentrations of reactants and products become constant.

Equilibrium constant (Kc): 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.

Balancing of equilibrium reaction: The many of chemical reaction involving for, reversible reaction is occurring backwards and forwards at the time by the same amount, it is balancing point of a chemical reaction, it seems to stop happening when the rates are equal equilibrium has occurred.

Equilibrium position: A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain spontaneous in neither direction.

Temperature affect in equilibrium: This process chemical shifts changes (or) towards the product or reactant, which can be determined by studying the reaction and deciding whether it is exothermic or endothermic.

Answer to Problem 15.4WE

The equilibrium constant (Kc) values are given the equilibrium reactions are shown below at 100⁰C.

$\begin{array}{l}(a).\text{K}c=\frac{{[NOBr]}^2}{{[NO]}^2[B{r}_2]}=71(b).\text{K}c={\left(\frac{{[NO]}^2[B{r}_2]}{{[NOBr]}^2}\right)}^2=2.0\times {10}^{-4}\\ (c).\text{K}c={\left(\frac{{[NO]}^2[B{r}_2]}{{[NOBr]}^2}\right)}^{\frac{1}{2}}=0.12(d).\text{K}c=\frac{{[NO]}^2{[BrCl]}^2}{{[NOBr]}^2[C{l}_2]}=0.10\\ (e).\text{K}c={\left(\frac{{[NOBr]}^2[C{l}_2]}{{[NO]}^2{[BrCl]}^2}\right)}^{1/2}=3.2\end{array}$

Explanation of Solution

To find: The equilibrium constant (Kc) values should be solved given the equilibrium reactions (a-e).

Write and analyze the equilibrium reactions fallowing model reactions (1 and 2).

The equilibrium constant (Kc) equation is given equilibrium reactions (1 and 2) shown below. Then determine the each relationship of each equations equilibrium expression to the equilibrium expression of the original equations and make the corresponding changes to the equilibrium constant for ach.

$\begin{array}{l}1).2\text{NOBr(g)}\rightleftharpoons \text{2NO(g)}\text{+}{\text{Br}}_{\text{2}}(g)----Kc=0.014\\ \text{Kc= }\frac{{[NO]}^2[B{r}_2]}{{[NOBr]}^2}\\ 2).B{r}_2+{\text{Cl}}_{\text{2}}(g)\rightleftharpoons 2BrCl(g)----Kc=7.2\\ \text{Kc= }\frac{{[BrCl]}^2}{[B{r}_2][C{l}_2]}\end{array}$

The equilibrium expression equation is showed above.

The equilibrium constant (Kc) value is solved given equilibrium reaction (a). Write and analyze the equilibrium reactions fallowing reactions.

To write the equilibrium equation (K) is the reverse of original equation (1). Its equilibrium expression is the reciprocal of that the original equation.

$\begin{array}{l}a).2\text{NO(g)}\text{+}{\text{Br}}_{\text{2}}(g)\rightleftharpoons \text{2NO}\text{Br}(g)\\ \text{Kc= }\frac{{[NOBr]}^2}{{[NO]}^2[B{r}_2]}=\frac{1}{0.014}=71\end{array}$

Reaction (b), to calculate the equilibrium constant value of above reaction (b),

$\begin{array}{l}b).4\text{NOBr(g)}\rightleftharpoons 4\text{NO}\text{+}2{\text{Br}}_{\text{2}}(g)\\ \text{Kc= }\left(\frac{{[NO]}^2[B{r}_2]}{{[NOBr]}^2}\right)\\ \text{Kc}={\left(\frac{1}{{\text{Kc}}^{\text{1}}}\right)}^2(or)=\left(\frac{1}{{{\text{(Kc}}^{\text{1}}\text{)}}^{\text{2}}}\right)=\frac{1}{71}=0.0140\\ \text{The respactive reaction (1) Kc values are substituted above equation}\\ ={(}^{0.014}=2.0\times {10}^{-4}\end{array}$

Above the original equation (1) multiplied by a factor of (2), its equilibrium expression is the original expression squared. The respective equilibrium constant value is showed above.

To calculate the equilibrium constant value of above reaction (c),

$\begin{array}{l}c).\text{NOBr(g)}\rightleftharpoons \text{NO}\text{+}\frac{1}{2}{\text{Br}}_{\text{2}}(g)\\ \text{Kc= }\sqrt{\frac{{[NO]}^2[B{r}_2]}{{[NOBr]}^2}}(or)=\left(\frac{{[NO]}^2[B{r}_2]}{{[NOBr]}^2}\right)\\ Kc={(}^{0.014}=0.12\end{array}$

Above the method is original equation (1) multiplied by. Further equilibrium expression is the square root of the original value.

To calculate the equilibrium constant value of above reaction (d)

$\begin{array}{l}d).2{\text{NOBr(g)+ Cl}}_{\text{2}}\rightleftharpoons 2\text{NO(g)}\text{+}2\text{BrCl}(g)\\ \text{Kc= }\frac{{[NO]}^2{[BrCl]}^2}{{[NOBr]}^2[C{l}_2]}\\ \text{The respactive reaction (1 and 2) Kc values are substituted above equation}\\ \text{Kc=(0}\text{.014)(7}\text{.2)}\\ Kc=0.10\end{array}$

 This is the sum of original equations (1 and 2). Its equilibrium expression is the product of the two individual expression, the solved (Kc) values are showed above.

Reaction (e): To calculate the equilibrium constant value of above reaction (e)

$\begin{array}{l}e).\text{NO(g)+ BrCl}(g)\rightleftharpoons \text{NOBr(g)}\text{+}\frac{1}{2}\text{Cl}(g)\\ \text{Kc= }\sqrt{\frac{{[NOBr]}^2[C{l}_2]}{{[NO]}^2[BrCl]}}(or){\left(\frac{{[NOBr]}^2[C{l}_2]}{{[NO]}^2[BrCl]}\right)}^{\frac{1}{2}}\\ \text{The respactive reaction (1 and 2) Kc values are substituted above equation}\\ \text{Kc=}{\left(\frac{1}{0.10}\right)}^{1/2}\\ Kc=3.2\end{array}$

Above the equation simplest way to analyze this reaction is to recognize that is the reverse of the reaction in reaction (d) multiplied by (1/2). Its equilibrium expression is the square root of the reciprocal of the expression in reaction (d).

The given equilibrium constant (Kc) equation is written by multiplying the activities for the species of the products and dividing by the activities of the reactants. Further uses the law of mass action to write the equilibrium expression for each equilibrium reaction, above these reactions are only gases and aqueous species appear in the expression. If any component in the reaction has a coefficient, indicated above with lower case letters, the concentration is raised to the power of the coefficient.

Further the each equilibrium constant will bears the same relationship to the original equation expression bears to the original.

Conclusion

The equilibrium constant (Kc) values are solved given the different set of equilibrium reactions.