Q22.73CP - Chemistry: Molecular Nature Of Matter And Change

Question

Nitric oxide occurs in the tropospheric nitrogen cycle, but it destroys ozone in the stratosphere. 

(a) Write equations for its reaction with ozone and for the reverse reaction. 

(b) Given tha$\mathbf{∆}{\mathit{G}}^{\mathbf{°}}$t the forward and reverse steps are first order in each component, write general rate laws for them. 

(c) Calculate for this reaction at$\mathbf{\text{280 K}}$ , the average temperature in the stratosphere. (Assume that the $\mathbf{∆}{\mathit{H}}^{\mathbf{°}}$and ${\mathit{S}}^{\mathbf{°}}$values in Appendix B do not change with temperature.) 

(d) What ratio of rate constants is consistent with$\mathit{K}$ at this temperature

Step-by-Step Solution

Answer

(a) The equations obtained are -

$$\begin{gathered} {\text{NO(g) + O}}_{\text{3}}\text{(g)}\to {\text{NO}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g)} \\ {\text{NO}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g)}\to {\text{NO(g) + O}}_{\text{3}}\text{(g)} \end{gathered}$$

(b) The general laws for forward and reverse steps are ${\text{Rate }}_{\text{For}}{\text{ = k}}_{\text{1}}·\text{[NO]}·\left[{\text{O}}_{\text{3}}\right]$and ${\text{Rate }}_{\text{Rev }}{\text{ = k}}_{\text{ - 1}}·\left[{\text{NO}}_{\text{2}}\right]·\left[{\text{O}}_{\text{2}}\right]$${\text{∆G}}^{°}\text{= - 198.8104 KJ}$respectively.

(c) The Gibbs Energy for the reaction is .$∆G^{°} = - 198.8104 KJ$

(d) The ratio of rate constants is ${\text{K = 1.23×10}}^{\text{37}}$.

Step 1: Concept Introduction

The nitrogen cycle is a biogeochemical cycle in which nitrogen is transformed into numerous chemical forms as it moves through the atmospheric, land, and sea ecosystems.

Step 2: Reactions with Ozone

(a)

In the tropospheric nitrogen cycle, the nitric oxide ($\text{NO}$) can react with ozone molecules, thus destroying the ozone layer as –

${\text{NO(g) + O}}_{\text{3}}\text{(g)}\to {\text{NO}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g)}$

However, the reverse reaction of ozone formation may also take place as the nitrogen dioxide- ${\text{NO}}_{\text{2}}$upon radiation, when striked with an energy of a wavelength, $295 \mu m<\lambda < 430 \mu m$, decompose into –

$$\begin{gathered} {\text{NO}}_{\text{2}}\text{(g)}\to \text{NO(g) + O(g)} \\ {\text{O(g) + O}}_{\text{2}}\text{(g)}\to {\text{O}}_{\text{3}}\text{(g)} \end{gathered}$$

Whereas, the overall equation for the reverse reaction can be written as –

${\text{NO}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g)}\to {\text{NO(g) + O}}_{\text{3}}\text{(g)}$

Therefore, the reactions obtained are

.$$\begin{gathered} {\text{NO(g) + O}}_{\text{3}}\text{(g)}\to {\text{NO}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g)}\backslash \mathrm{hfill} \\ {\text{NO}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g)}\to {\text{NO(g) + O}}_{\text{3}}\text{(g)}\backslash \mathrm{hfill} \end{gathered}$$

Step 3: General Rate Laws

(b)

The general rate law then, for the forward reaction if it is first order for each component is then –

${\text{Rate }}_{\text{For}}{\text{ = k}}_{\text{1}}·\text{[NO]}·\left[{\text{O}}_{\text{3}}\right]$

The reverse reaction general rate then –

${\text{Rate }}_{\text{Rev }}{\text{ = k}}_{\text{ - 1}}·\left[{\text{NO}}_{\text{2}}\right]·\left[{\text{O}}_{\text{2}}\right]$

Therefore, the general rate laws are${\text{Rate }}_{\text{For}}{\text{ = k}}_{\text{1}}·\text{[NO]}·\left[{\text{O}}_{\text{3}}\right]$ and ${\text{Rate }}_{\text{Rev }}{\text{ = k}}_{\text{ - 1}}·\left[{\text{NO}}_{\text{2}}\right]·\left[{\text{O}}_{\text{2}}\right]$.

Step 4: Calculation for Gibbs Energy

(c)

For the forward reaction, Gibb's energy at is calculated upon calculation of the enthalpy and entropy ofthe reaction such that –

$$\begin{gathered} ∆H_{rxn}=\sum ∆H_{products}--\sum ∆H_{reac\tan ts} \\ ∆H_{rxn}=\left[1 mol\times ∆{\text{H(NO}}_2\text{)+1 mol×∆H}\left({\text{O}}_2\right)\right]-\left[1 mol\times ∆\text{H(NO)+1 mol×∆H}\left({\text{O}}_3\right)\right] \\ ∆H_{rxn}=\left[1 mol\times 33.20\text{kJ/mol+ 1 mol×0.00kJ/mol}\right]-\left[1 mol\times 90.29\text{kJ/mol+ 1 mol×143.00kJ/mol}\right] \\ ∆H_{rxn}=\text{ - 200.09KJ} \end{gathered}$$

$$\begin{gathered} ∆S_{rxn}=\left[1 mol\times ∆{\text{S(NO}}_2{\text{)+ 1 mol×S(O}}_2\text{)} \right]-\left[1 mol\times ∆{\text{S(NO)+ 1 mol×S(O}}_3\text{)} \right] \\ ∆S_{rxn}=\left[1 mol\times \text{239.90}\frac{J}{mol\times k}\text{+}1 mol\times \text{205.00}\frac{{\displaystyle J}}{{\displaystyle mol\times k}}\right]-\left[1 mol\times \text{210.65}\frac{J}{mol\times k}\text{+}1 mol\times 238.82\frac{{\displaystyle J}}{{\displaystyle mol\times k}}\right] \end{gathered}$$

$$\begin{gathered} {\text{∆S}}_{rxn}\text{ = - 4.57}\frac{\text{J}}{\text{K}} \\ \text{ = - 4.57}·{\text{10}}^{\text{ - 3}}\frac{\text{kJ}}{\text{K}} \end{gathered}$$

Assuming that enthalpy and entropy change is not significant with temperature, at  $280k$–

$$\begin{gathered} ∆G_{25}=-\text{ - 200.09KJ - (280)K×}\left({\text{ - 4.57×10}}^{\text{ - 3}}\right)\frac{\text{kJ}}{\text{K}} \\ ∆G_{25}\text{= - 198.8104KJ} \end{gathered}$$

Therefore, the value for Gibbs Energy is obtained as$- 198.8104\mathrm{KJ}$ .

Step 5: Ratio of Rate Constant

(d)

At equilibrium, the forward reaction rate is equal to the rate of the reverse reaction. Thus, it is obtained –

$$\begin{gathered} {\text{Rate }}_{\text{For}}{\text{ = Rate }}_{\text{Rev}} \\ {\text{k}}_{\text{1}}·\text{[NO]}·\left[{\text{O}}_{\text{3}}\right]{\text{ = k}}_{\text{ - 1}}·\left[{\text{NO}}_{\text{2}}\right]·\left[{\text{O}}_{\text{2}}\right] \\ \frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{ - 1}}}\text{ = }\frac{\left[{\text{NO}}_{\text{2}}\right]·\left[{\text{O}}_{\text{2}}\right]}{\text{[NO]}·\left[{\text{O}}_{\text{3}}\right]} \end{gathered}$$

The ratio of rate constants is defined asaequilibrium constant, $k$ –

$\text{K = }\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{ - 1}}}$

From$∆G$ , the $k$can be expressed as –

$$\begin{gathered} \text{∆G = - R}·\text{T}·\text{lnK} \\ \text{lnK = }\frac{\text{∆G}}{\text{ - R}·\text{T}} \\ {\text{K = e}}^{\text{lnK}} \end{gathered}$$

Hence, from previous part it is obtained that –

$$\begin{gathered} \text{lnK =}\frac{-198.8104\times {10}^3{\displaystyle \text{J}}{\displaystyle ·}{\displaystyle \text{mol}}}{{\displaystyle \text{ - 8.314}\frac{\text{J}}{\text{mol}·\text{K}}\text{×280 K}}} \\ \text{lnK = 85.40259 } \\ {\text{K = 1.23×10}}^{\text{37}} \end{gathered}$$

Therefore, the value for ratio is obtained as${\text{K = 1.23×10}}^{\text{37}}$${\text{K = 1.23×10}}^{\text{37}}$ .