The thermodynamic stability of NO(g) based on the given data is:

$\frac{1}{2}{N_2}_{<mspace\; width="1em"></mspace>\left(\mathrm{g}\right)}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{1}{2}{O}_{2<mspace\; width="1em"></mspace>\left(\mathrm{g}\right)}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>N{O}_{\left(\mathrm{g}\right)}<mspace\; width="1em"></mspace>;$
${∆}_{\mathrm{r}}H°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>90<mspace\; width="1em"></mspace>kJ<mspace\; width="1em"></mspace>mo{l}^{-1}$
$N{O}_{\mathit{\left(}g\mathit{\right)}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{1}{2}{O_2}_{\mathit{\left(}g\mathit{\right)}}<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>N{O_{\mathit{2}}}_{\mathit{\left(}g\mathit{\right)}}\hspace{0.17em};<mspace\; width="1em"></mspace>$
${∆}_{\mathrm{r}}\mathrm{H}°<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>-74<mspace\; width="1em"></mspace>\mathrm{kJ}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$

1. Less than NO_2(g) 

2. More than NO_2(g)

3. Equal to NO_2(g)

4. Insufficient data

The entropy change in the surroundings when 1.00 mol of H₂O_(l) is formed under standard conditions is:

[Given: ∆_fH^θ = –286 kJ mol^–1 ]

1. 952. 5 J mol^-1

2. $979.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

3. $949.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

4. $959.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

Determine the value of  ∆G°, if the equilibrium constant for a reaction is 10:

($\mathrm{R}$ $=$ $8.314$ $\mathrm{J}$ ${\mathrm{K}}^{-1}$ ${\mathrm{mol}}^{-1}$ $;$ $\mathrm{T}$ $=$ $300$ $\mathrm{K}$)

1. \(-5.74 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(-5.74 \mathrm{~J} \mathrm{~mol}^{-1}\)
3. \( +4.57 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(-57.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The value of ∆G°  for the given reaction would be:
\( 2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow 2 \mathrm{D}(\mathrm{~g})\)
(Given: ∆U° = – 10.5 kJ and  ∆S° = – 44.1 J K^–1)

For the reaction:

The enthalpy change (ΔH) is 400 kJ mol⁻¹ and entropy change (ΔS) is 0.2 kJ K⁻¹ mol⁻¹.

Determine the temperature at which the reaction becomes spontaneous:

For an isolated system with ∆U = 0, the ∆S value will be:

${∆}_{\mathrm{vap}}\mathrm{H}°$ $\left({\mathrm{CCl}}_4\right)$ $=$ $30.5$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${∆}_{\mathrm{f}}\mathrm{H}°$ $\left({\mathrm{CCl}}_4\right)$ $=$ $-$ $135.5$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${∆}_{\mathrm{a}}\mathrm{H}°$ $\left(\mathrm{C}\right)$ $=$ $715.0$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${∆}_{\mathrm{a}}\mathrm{H}°$ $\left({\mathrm{Cl}}_2\right)\hspace{0.17em}=$ $242$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

The enthalpy change for the following reaction would be:

 ${\mathrm{CCl}}_4$ $\left(\mathrm{g}\right)$ $\to$ $\mathrm{C}\left(\mathrm{g}\right)$ $+$ $4\mathrm{Cl}$ $\left(\mathrm{g}\right)$ 

The standard enthalpy of the formation of CH₃OH_(l) from the following data is:

For the reaction:
N₂(g) + 3H₂(g) → 2NH₃(g) ; ΔᵣH° = –92.4 kJ mol⁻¹

What is the standard enthalpy of formation (ΔfH°) of NH₃(g)?

The enthalpy of formation of ${\mathrm{CO}}_{\left(\mathrm{g}\right)},$ ${\mathrm{CO}}_{2\left(\mathrm{g}\right)},$ ${\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)},\hspace{0.17em}\mathrm{and}$ ${\mathrm{N}}_2{\mathrm{O}}_{4\left(\mathrm{g}\right)}$ $$are –110 kJ ${\mathrm{mol}}^{-1}$, – 393 kJ ${\mathrm{mol}}^{-1}$, 81 kJ ${\mathrm{mol}}^{-\mathrm{1,}}$ and 9.7 kJ \(\text{mol}^{- 1}\) respectively. The value of \(\left(\Delta\right)_{r} H\) for the following reaction would be:

\(\mathrm{N_{2} O_{4 \left(g\right)} + 3 CO{\left(g\right)} \rightarrow N_{2} O_{\left(g\right)} + 3CO_{2 \left(g\right)}}\)