Chapter 17, Problem 17.104P

(a)

Interpretation Introduction

Interpretation:

${\text{K}}_{\text{p}}$ for the given reaction $\text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)$ has to be calculated.

Concept Introduction:

The relation between ${\text{K}}_{\text{p}}$ and ${\text{K}}_{\text{c}}$ is given by the following equation.

${\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\Delta {\text{n}}_{\text{gas}}}\boxed{\begin{array}{l}{\text{K}}_{\text{p}}\text{ = }Equilibriumcons\tan tintermsofpartialpressure\\ {\text{K}}_{\text{c}}\text{ = }Equilibriumcons\tan tintermsofconcentration\\ \Delta {\text{n}}_{\text{gas}}\text{ = moles of gaseous product }-\text{ moles of gaseous reactant}\end{array}}$

Only moles of gaseous products and reactants are used for calculating $\Delta {\text{n}}_{\text{gas}}.$

Answer to Problem 17.104P

${\text{K}}_{\text{p}}$ for the given reaction is $4.9\times 10^5.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}\text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)\\ \\ {\text{K}}_{\text{p}}\text{ = 1}\text{.3}\times {\text{10}}^4\\ \\ T\text{ = 457 K}\end{array}$

• Determine $\Delta n_{gas}$:

The total number of moles of gaseous reactants is 3 and the moles of gaseous product is 2.  $\Delta n_{gas}$ is determined as follows,

  $\begin{array}{l}\Delta {\text{n}}_{\text{gas}}\text{= moles of gaseous product - moles of gaseous reactant}\\ =2-3\\ =-1\end{array}$

• Determine ${\text{K}}_{\text{p}}$:

${\text{K}}_{\text{p}}$ of the reaction can be determined from given ${\text{K}}_{\text{c}}$ as below,

  $\begin{array}{c}{\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\Delta {\text{n}}_{\text{gas}}}\\ \\ =\left(1.3\times {10}^4\right){\left[\left(0.0821\text{ atm}\text{.L/mol}\text{.K}\right)\left(457\text{ K}\right)\right]}^{-1}\\ \\ =\text{ 4}\text{.9}\times {\text{10}}^5\end{array}$

Therefore,

${\text{K}}_{\text{p}}$ for the given reaction is $4.9\times 10^5.$

(b)

Interpretation Introduction

Interpretation:

Standard heat of formation for the given reaction , $\text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)$ has to be calculated.

Concept Introduction:

The value of standard heat of formation ${\text{ΔH}}^{\text{ο}}$ of the given reaction is calculated by the formula,

${\text{ΔH}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

$\begin{array}{c}{\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}\text{=}\text{the standard enthalpy of formation for the reactants}\\ {\text{ΔH}}_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}\text{=}\text{the standard enthalpy of formation for the products}\\ {\text{n}}_{\text{p}}\text{=}\text{number of products molecule}\\ {\text{n}}_{\text{r}}\text{=}\text{number of reactants molecule}\end{array}$

Answer to Problem 17.104P

Standard heat of formation for the given reaction is $-114.2kJ.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}\text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)\\ \\ {\text{ΔH}}_f^o\left[N{O}_2\left(g\right)\right]\text{ = 33}\text{.2 kJ/mol}\\ \\ {\text{ΔH}}_f^o\left[NO\left(g\right)\right]\text{ = 90}\text{.29 kJ/mol}\\ \\ {\text{ΔH}}_f^o\left[O_2\left(g\right)\right]\text{ = 0 kJ/mol}\end{array}$

Standard heat of formation for the given reaction can be determined given below,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \\ =\left(2\times {\text{ΔH}}_f^o\left[N{O}_2\left(g\right)\right]\right)-\left[\left(2\times {\text{ΔH}}_f^o\left[NO\left(g\right)\right]\right)+{\text{ΔH}}_f^o\left[O_2\left(g\right)\right]\right]\\ \\ =\left(2\text{ mol}\times \text{33}\text{.2 kJ/mol}\right)-\left[\left(2\text{ mol}\times \text{90}\text{.29 kJ/mol}\right)+\left(1\text{ mol}\times \text{0 kJ/mol}\right)\right]\\ \\ =-\text{114}\text{.2 kJ}\end{array}$

Therefore,

Standard heat of formation for the given reaction is $-114.2kJ.$

(c)

Interpretation Introduction

Interpretation:

Temperature at which $K_c=\text{ 6}\text{.4}\times {10}^9$ for the reaction $\text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)$ has to be calculated.

Concept Introduction:

The van’t Hoff Equation: Effect of temperature on equilibrium constant can be shown as follows,

  $\begin{array}{c}\ln \frac{K_2}{K_1}=-\frac{\Delta H_{rxn}^o}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\\ \\ where,\\ {\text{K}}_1{\text{ = Equilibrium constant at T}}_{\text{1}}\\ {\text{K}}_2{\text{ = Equilibrium constant at T}}_2\\ \Delta H_{rxn}^o=S\tan dardheatofformationforthegivenreaction\end{array}$

Answer to Problem 17.104P

Standard heat of formation for the given reaction is $-114.2kJ.$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}\text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)\\ \\ {\text{ΔH}}_{rxn}^o\text{ = }-\text{114}\text{.2 kJ}\\ \\ T_1\text{ = 457 K}\\ \\ {\text{K}}_1\text{ at 457 K = 4}\text{.9}\times {\text{10}}^5\\ \\ K_2=\text{ 6}\text{.4}\times {10}^9\end{array}$

Temperature at which $K_c=\text{ 6}\text{.4}\times {10}^9$ for given reaction can be determined given below,

  $\begin{array}{l}\ln \frac{K_2}{K_1}=-\frac{\Delta H_{rxn}^o}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\\ \\ \ln \frac{\text{6}\text{.4}\times {10}^9}{\text{4}\text{.9}\times {\text{10}}^5}=-\frac{-\text{114}\text{.2 kJ}}{\left(\text{8}\text{.314 J/mol K}\right)}\left(\frac{1}{T_2}-\frac{1}{\text{457 K}}\right)\\ \\ T_2\text{ = 347}\text{.4 K}\end{array}$

Therefore,

Temperature at which $K_c=\text{ 6}\text{.4}\times {10}^9$ for given reaction is $347.4K.$