91Ó°ÊÓ

The thermodynamics of each step - i.e., Gibb's energy of each reaction step can be analyzed to solve the question.

${\text{N}}_2\left(g\right)+2{\text{O}}_2\left(g\right)→2{\text{NO}}_2\left(g\right)$

Calculate the reaction's enthalpy and entropy using Appendixfor reaction (1).

role="math" localid="1663389271652" $$\begin{gathered} ∆H_{rxn}=∑∆H_{Products}-∑∆H_{reactants} \\ ∆H_{rxn}=\left[2\text{mol}·\mathrm{Δ}H\left(N{O}_2\right)\right]-\left[1\text{mol}·\mathrm{Δ}H\left(N_2\right)+2\text{mol}·\mathrm{Δ}H\left(O_2\right)\right] \\ ∆H_{rxn}=[2\text{mol}·33.20\text{kJ}/\text{mol}]-[1\text{mol}·0.00\text{kJ}/\text{mol}+2\text{mol}·0.00\text{kJ}/\text{mol}] \\ ∆H_{rxn}=\mathbf{66}.\mathbf{40}\text{kJ} \\ ∆s_{rxn}=∑s_{Products}-∑s_{mol}=∆S\left(N{S}_2\right)]-\left[1\text{mol}·∆S\left(N_2\right)+2\text{mol}·∆S\left(O_2\right)\right] \\ ∆s_{rxn}=\left[2\text{mol}·239.90\frac{\text{J}}{\text{mol}·\text{K}}\right]-\left[1\text{mol}·191.50\frac{\text{J}}{\text{mol}·\text{K}}+2\text{mol}·205.00\frac{\text{m}}{\text{mol}·\text{K}}\right] \end{gathered}$$

$$\begin{gathered} ∆s_{rxn}=-121.70\frac{\text{J}}{\text{K}} \\ =-0.1217\frac{\text{kJ}}{\text{K}} \end{gathered}$$

The reaction 1 is nonspontaneous at all temperatures because it is endothermic $(∆H>0)$ and the overall entropy of the reaction drops. As Gibb's energy is positive, no spontaneous reaction occurs.

$$\begin{gathered} ∆G_{25}=66.40\text{kJ}-(25+273)\text{K}·-0.1217\frac{\text{kJ}}{\text{K}} \\ ∆G_{25}=102.67\text{kJ} \end{gathered}$$

Hence, it represents non-spontaneous reactions at all temperatures, so use of catalyst

$3{\text{NO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)→2{\text{HNO}}_3\left(aq\right)+\text{NO}\left(g\right)$

Calculate the reaction's enthalpy and entropy using Appendix B for reaction (2).

$$\begin{gathered} ∆H_{rxn}=\left[2\text{mol}·\mathrm{Δ}H\left(HN{O}_3\right)+1\text{mol}·\mathrm{Δ}H\left(NO\right)\right] \\ -\left[3\text{mol}·\mathrm{Δ}H\left(N{O}_2\right)+2\text{mol}·\mathrm{Δ}H\left(H_{2}O\right)\right] \end{gathered}$$

$$\begin{gathered} ∆H_{rxn}=[2\text{mol}·-206.57\text{kJ}/\text{mol}+1\text{mol}·90.29\text{kJ}/\text{mol}] \\ -[3\text{mol}·33.20\text{kJ}/\text{mol}+2\text{mol}·-285.83\text{kJ}/\text{mol}] \\ ∆H_{rxn}=\mathbf{149}.\mathbf{21}\text{kJ} \\ ∆s_{rxn}=\left[2\text{mol}·\mathrm{Δ}S\left(HN{O}_3\right)+1\text{mol}·\mathrm{Δ}S\left(NO\right)\right] \\ -\left[3\text{mol}·\mathrm{Δ}S\left(N{O}_2\right)+2\text{mol}·\mathrm{Δ}S\left({\text{H}}_{2}O\right)\right] \end{gathered}$$

$$\begin{gathered} ∆s_{rxn}=\left[2\text{mol}·146.00\frac{\text{J}}{\text{mol}·\text{K}}+1\text{mol}·210.65\frac{\text{J}}{\text{mol}·\text{K}}\right] \\ -\left[3\text{mol}·239.90\frac{\text{J}}{\text{mol}·\text{K}}+2\text{mol}·69.95\frac{\text{J}}{\text{mol}·\text{K}}\right] \end{gathered}$$

$$\begin{gathered} ∆s_{rxn}=-356.95\frac{\text{J}}{\text{K}} \\ =-0.35695\frac{\text{kJ}}{\text{K}} \end{gathered}$$

The reaction 2 is nonspontaneous at all temperatures because it is endothermic$\left(DeltaH>0\right)$and the overall entropy of the reaction drops. For example, Gibb's energy is positive at${25}^{°}C$,indicating that the reaction is not spontaneous.

localid="1663390646989" $$\begin{gathered} ∆G_{25}=149.21\text{kJ}-\left\{(25+273)\text{K}×-0.35695\frac{\text{kJ}}{\text{K}}\right\} \\ ∆G_{25}=255.581\text{kJ} \end{gathered}$$

Therefore, it represents non-spontaneous reactions at all temperatures, so use of catalyst is required.

$2\text{NO}\left(g\right)+{\text{O}}_2\left(g\right)→2{\text{NO}}_2\left(g\right)$

Calculate the reaction's enthalpy and entropy using Appendix B for reaction (2).

$$\begin{gathered} ∆H_{rxn}=\left[2\text{mol}·\mathrm{Δ}H\left({\text{NO}}_2\right)\right]-\left[2\text{mol}·\mathrm{Δ}H\left(NO\right)+1\text{mol}·\mathrm{Δ}H\left(O_2\right)\right] \\ ∆H_{rxn}=[2\text{mol}·33.20\text{kJ}/\text{mol}]-[2\text{mol}·90.29\text{kJ}/\text{mol}+1\text{mol}·0.00\text{kJ}/\text{mol}] \\ ∆S_{rxn}=\left[2\text{mol}·∆S\left(N{\text{H}}_{\text{rxn}}=-\mathbf{114}.\mathbf{18}\text{kJ}-\left[2\text{mol}·∆S\left(\text{mO}\right)+1\text{mol}·∆S\left(O_2\right)\right]\right.\right. \\ ∆S_{rxn}=\left[2\text{mol}·239.90\frac{\text{J}}{\text{mol}·\text{K}}\right]-\left[2\text{mol}·210.65\frac{\text{J}}{\text{mol}·\text{K}}+1\text{mol}·205.00\frac{\text{J}}{\text{mol}·\text{K}}\right] \\ {\mathbf{Δ³\S }}_{\text{rxn}}=-\mathbf{146}.\mathbf{50}\frac{\text{J}}{\text{K}}=-\mathbf{0}.\mathbf{1465}\frac{\text{kJ}}{\text{K}} \end{gathered}$$

At low temperatures, the reaction 3 is spontaneous because it is exothermic$\left(DeltaH>0\right)$and the overall entropy of the process reduces. For instance, for $${25}^{°}C$,, Because Gibb's energy is positive, the reaction occurs naturally.

$∆G_{25}=-114.18\text{kJ}-(25+273)\text{K}·-0.1465\frac{\text{kJ}}{\text{K}}$

$∆G_{25}=-70.523\text{kJ}$

$∆G_{25}=-70.523\text{kJ}$

width="145">∆G25=-70.523kJ

As a result, the reaction of step (3) is spontaneous only at low temperatures.

As a result, the entire reaction is implausible, because the first two reactions are not spontaneous and only the third is.

In the industrial scale, preheated air is mixed with ammonia gas and acatalyst is employed to produce nitric oxide, which is then oxidized and, in the presence of water, nitric acid is formed.

At all temperatures, the reaction steps (1) and (2) indicate non-spontaneous reactions, necessitating the employment of a catalyst. On the other hand, Step (3), is only spontaneous at low temperatures.As a result, the entire reaction is implausible