Chapter 15, Problem 15.95QP

Interpretation Introduction

Interpretation:

To Calculate and analyze the each partial pressure $\left(\text{Kp}\right)$ values of given N2O4 into NO2 equilibrium reaction with respective temperature at ${25}^{\text{ο}}\text{C}$.

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.

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. 

Kp and Kc: This equilibrium constants of gaseous mixtures, these difference between the two constants is that Kc is defined by molar concentrations, whereas Kp is defined by the partial pressures of the gasses inside a closed system.

Answer to Problem 15.95QP

The reactant and product equilibrium (Kp) pressure values are given the statement of (NO₂) homogenous reaction is shown below.

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2NO}}_{\text{2}}{}_{\text{(g) }}\\ {\text{K}}_p=\frac{{\text{(P}}_{{\text{NO}}_{\text{2}}})}{(P_{N_2{O}_4})}=0.113\\ \text{a)}\text{.}{P}_{N_2{O}_4}=0.09atm\\ \text{b)}\text{.}{\text{P}}_{\text{total}}=0.10\text{atm}\end{array}$

Explanation of Solution

To find: Calculate the each partial pressure $(Kc)$ values for given the statement of NO2 and N2O4 equilibrium reaction.

Calculate and analyze the $(Kp)$ values with respective statement.

First we derived the equilibrium pressure values of given statement

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2NO}}_{\text{2}}{}_{\text{(g) }}\\ {\text{K}}_p=\frac{{\text{(P}}_{{\text{NO}}_{\text{2}}})}{(P_{N_2{O}_4})}-------[1]\\ Giventheequilibriumvaluesare\\ N{O}_2=0.15\text{atm}\text{and}\text{N2O4=0}\text{.20}atm\\ Thisvaluesaresubstitutedequation(1)\\ {\text{K}}_p=\frac{{(0.15)}^2}{(0.20)}=0.113\end{array}$

Here volume is double so pressure is halved, further the calculate the reaction quintet (Qp) and compare it to Kp.

  $\begin{array}{l}Qp=\frac{{\left(\frac{0.15}{2}\right)}^2}{\left(\frac{0.20}{2}\right)}\\ =0.0563\angle Kp\end{array}$

So the equilibrium will shifted to the right, some ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ will react some ${\text{NO}}_{\text{2}}$ will formed, hence (x) amount of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ reacted.

Hence we derive here equilibrium pressure values of reactant and product.

  $\begin{array}{l}\text{Hare, }\\ {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}\rightleftharpoons {\text{2NO}}_{\text{2}}\text{(g) }\\ \text{Initial (atm): }0.100.075\\ \text{Change (atm): }\text{-x}+2x\\ ---------------------\\ \text{Eqilibrium (atm):}0.10-x0.075+2x\\ \text{Solving for the equilibrium constant: }\\ {\text{K}}_p=\frac{{\text{(P}}_{{\text{NO}}_{\text{2}}})}{(P_{N_2{O}_4})}-------[1]\\ \text{The equilibrium pressure values are substituted above the equation (1)}\\ \text{Kp=}0.113\\ \text{Kp=}\frac{{(0.075+2x)}^2}{(0.10-x)}\\ Kp=0.113atm\\ Hence0.113=\frac{{(0.075+2x)}^2}{(0.10-x)}------[2]\\ Solvedaboveequation(2)\\ 4x^2+0.413-5.68\times {10}^{-3}=0\\ x=0.0123\\ \end{array}$

Finally we solved the both equilibrium as fallowed as

  $\begin{array}{l}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(g)}\rightleftharpoons {\text{2NO}}_{\text{2}}\text{(g) }\\ {\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}=0.10-x\\ Herex=0.0123\\ {\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}=0.10-0.0123=0.09atm\\ {\text{P}}_{{\text{NO}}_{\text{2}}}=0.0745+2x\\ x=0.0123\\ {\text{P}}_{{\text{NO}}_{\text{2}}}=0.0745+2(0.0123)\\ {\text{P}}_{{\text{NO}}_{\text{2}}}=0.0996\text{taking}\text{for}\text{rounded}\text{values}\\ {\text{P}}_{{\text{NO}}_{\text{2}}}\approx 0.10\text{atm}\end{array}$

The given ${\text{NO}}_{\text{2}}$ formation reaction the respective reactant to give products (one products) all exists in the different phase and this equilibrium reaction expression contains different conditions like solid converted into gases phase, so this equilibrium reactions has heterogeneous.   The equilibrium constant can also be represented by Kp, were the “P” partial pressure. The each partial pressure, total and degree of dissociation values are derived given ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ into ${\text{NO}}_{\text{2}}$ equilibrium reaction equation at ${25}^{\text{0}}\text{C}$ as showed above.

Conclusion

The each reactant and product values are calculated given the nitrogen dioxide formation reaction and corresponding temperature at ${25}^{\text{0}}\text{C}$.