Q93CP - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the following reaction:

$\mathbf{\text{2NOBr(g) }}\mathbf{\rightleftharpoons }{\mathbf{\text{2NO(g) + Br}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{ }$

  $\mathit{K}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{42}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathbf{373}\mathit{K}$

Given that  ${\mathit{S}}^{\mathbf{°}}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{N}\mathit{O}\mathit{B}\mathit{r}\left(g\right)\mathbf{=}\mathbf{272}\mathbf{.}\mathbf{6}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{\times }\mathit{K}$  and that$\mathbf{∆}{\mathit{S}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathbf{∆}{\mathit{H}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}$   are constant with temperature, find

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{∆}{\mathit{S}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathbf{298}\mathit{K} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{∆}{\mathit{G}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathbf{373}\mathit{K} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{ }\mathbf{∆}{\mathit{H}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathbf{373}\mathit{K} \\ \mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{ }\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{°}}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{N}\mathit{O}\mathit{B}\mathit{r}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathbf{298}\mathit{K} \\ \mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{∆}{\mathit{G}}_{\mathbf{\gamma }\mathbf{g}\mathbf{n}}^{\mathbf{°}}\mathit{a}\mathit{t}\mathbf{ }\mathbf{298}\mathit{K} \\ \mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{ }\mathbf{∆}{\mathit{G}}_{\mathbf{f}}^{\mathbf{°}}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{N}\mathit{O}\mathit{B}\mathit{r}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathit{a}\mathit{t}\mathbf{ }\mathbf{298}\mathit{K} \end{gathered}$$

 

Step-by-Step Solution

Answer

a) The entropy change of the reaction at$298K is 121.48\frac{J}{K\times mol}$  .

b)The Gibb's energy change of the reaction at $373 is 2.690\frac{kJ}{mol}$ .

c) The enthalpy of the reaction at $373K is 48.002\frac{kJ}{mol}$ .

d) The enthalpy of $NOBr\left(g\right) at 298K is 81.722\frac{kJ}{mol}$ .

e) The $∆G_{rxn}^{°} at 298 K is 11.801\frac{kJ}{mol}$ .

f) The$∆G_f^{°} at 298K is 82.265 \frac{kJ}{mol}$  .

Step 1: Concept Introduction

The chemical compound with the chemical formula   is nitrosyl bromide. It's a red gas that condenses slightly below room temperature. The reversible reaction of nitric oxide with bromine can produce nitrosyl bromide.

Step 2: Calculating ∆ S r x n °   a t   298   K

 To calculate the standard entropy of the reaction, use the values given in Appendix B.

$$\begin{gathered} ∆S_{rxn}^{°}=∆S_{product}^{°}-∆S_{reac\tan t}^{°}∆S_{rxn}^{°} \\ =\left[2∆S^{°}{\text{(NO(g))+1∆S}}^{°}\left({\text{Br}}_{\text{2}}\text{(g)}\right)\right]-\left[2∆S^{°}\text{(NOBr(g))}\right] \\ ∆S_{rxn}^{°}=\left[2mol\times \left(210.65\frac{J}{mol \times K}\right)+1mol\times \left(245.38\frac{J}{mol \times K}\right)\right]-\left[2mol\times \left(272.60\frac{J}{mol \times K}\right)\right] \\ ∆S_{rxn}^{°}=121.48\frac{J}{K} \end{gathered}$$

 

The entropy change of the reaction at $298K is 121.48\frac{J}{K}$  .

Step 3: Calculating ∆ G r x n °   a t   373   K

  Gibb's energy change can be defined as,

 $∆G_{rxn}^{°}=-R\times T\times InK$

Since the temperature and $K$  are known, the $∆G_{rxn}^{°}$  can be calculated at $373K$  as,

The $∆G_{rxn}^{°} at 298K is 11.801 kJ/mol$

Step 4: Calculating ∆ H r x n °   a t   373   K

  From the equation,

 $∆G_{Txn}^{°}=∆H_{Txn}^{°}-T\times ∆S_{Txn}^{°}$

The   $∆H_{rxn}^{°} at 373K$can be expressed, considering that entropy is constant with temperature:

$$\begin{gathered} ∆H_{rxn}^{°}=∆G_{rxn}+T\times ∆S_{rxn}^{°} \\ ∆H_{rxn}^{°}=2.690\frac{kJ}{mol}+373K\times \left(121.48\times {10}^{-3}\frac{kJ}{K}\right) \\ ∆H_{rxn}^{°}=48.002\frac{kJ}{mol} \end{gathered}$$ $$

The enthalpy of the reaction at $373K is 48.002\frac{kJ}{mol}$ .

Step 5: Calculating ∆ H f °   o f   N O B r   a t   298 K

d) The entropy of  $NOBr$ can be determined as follows: Since enthalpy is a state function, it can be represented as the difference between all products and reactants entropies:

 $$\begin{gathered} ∆H_{rxn}^{°}=∆H_{product}^{°}-∆H_{reac\tan t}^{°} \\ ∆H_{rxn}^{°}=\left[2∆H^{°}{\text{(NO(g))+1∆H}}^{°}\left({\text{Br}}_{\text{2}}\text{(g)}\right)\right]-\left[2∆H^{°}\text{(NOBr(g))}\right] \\ ∆H^{°}\text{(NOBr(g))=}\frac{1}{2}\text{×}\left[2∆H^{°}{\text{(NO(g))+1∆H}}^{°}\left({\text{Br}}_{\text{2}}\text{(g)}\right){\text{-∆H}}_{rxn}^{°}\right] \end{gathered}$$

Then, using Appendix B, and assuming that enthalpy is constant with temperature

  $$\begin{gathered} ∆H^{°}\text{(NOBr(g))=}\frac{1}{2}\text{×}\left[2mol\times 90.29\frac{kJ}{mol}+1mol\times 30.91\frac{kJ}{mol}-48.002\frac{kJ}{mol}\right] \\ {\text{∆H}}^{°}\text{(NOBr(g))=81.744kJ/mol} \end{gathered}$$

The enthalpy of $NOBr at 298K is 81.744\frac{kJ}{mol}$ .

Step 6: Calculating ∆ G γ g n °   a t   298 K

  Gibb's energy change can be defined as

  

Assuming that entropy and enthalpy are constant with temperature, at$298K$

$$\begin{gathered} ∆G_{rxn}^{°}= 48.002\frac{kJ}{mol}-298K\times \left(121.48\times {10}^{-3}\frac{J}{mol \times K}\right) \\ ∆G_{rxn}^{°}=11.80096\frac{kJ}{mol} \end{gathered}$$

  

 The  $∆G_{rxn}^{°} at 298 K is 11.801kJ/mol$

Step 7: Calculating ∆ G f °   o f   N O B r   a t   298 K

  Because $∆G_{Txn}^{°}$  is a state function that can be written as the difference between all products and reactants $∆G_{f^{°}}^{°} the∆G_f^{°} of the NOBr$  may be determined as follows:

 $$\begin{gathered} ∆G_{rxn}^{°}=∆G_{product}^{°}-∆G_{reac\tan t}^{°} \\ ∆G_{rxn}^{°}=\left[2∆G^{°}{\text{(NO(g))+ 1∆G}}^{°}\left({\text{Br}}_{\text{2}}\text{(g)}\right)\right]-\left[2∆G^{°}\text{(NOBr(g))}\right] \\ ∆G^{°}\text{(NOBr(g))=}\frac{1}{2}\text{×}\left[2∆G^{°}{\text{(NO(g))+1∆G}}^{°}\left({\text{Br}}_{\text{2}}\text{(g)}\right){\text{-∆G}}_{rxn}^{°}\right] \end{gathered}$$

Then, using Appendix B, and the value calculated previously in e)

 $$\begin{gathered} ∆G^{°}\text{(NOBr(g))=}\frac{1}{2}\text{×}\left[2 mol\times 86.60\frac{kJ}{mol}+1mol\times 3.13\frac{kJ}{mol}-11.801\frac{kJ}{mol}\right] \\ {\text{∆H}}^{°}\text{(NOBr(g))=82.2645kJ/mol} \end{gathered}$$

The   $∆G_f^{°} of NOBr at 298K is 82.265\frac{kJ}{mol}$.