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

For the graph given below, it can be concluded that work done during the process shown will be-

Consider the following graph.

The work done, as per the graph above, is:

Consider the following diagram for a reaction $\mathrm{A}\to \mathrm{C}$. 

The nature of the reaction is-

1. Exothermic

2. Endothermic

3. Reaction at equilibrium

4. None of the above

What is the nature of the reaction depicted in the given diagram for A→C?

An enthalpy diagram for a particular reaction is given below:

Which of the following statements is correct?

The difference between enthalpy (∆H) and internal energy (∆E) for the given below reaction,
under constant temperature, is:
\(C_{3} H_{8} \left(\right. g \left.\right) + 5 O_{2} \left(\right. g \left.\right) \rightarrow 3 CO_{2} \left(\right. g \left.\right) + 4 H_{2} O \left(\right. l \left.\right)\)

For the given reaction 

$2H_2{O}_2\left(l\right)$ $\to$ $2H_{2}O\left(l\right)$ $+$ $O_2\left(g\right)$, the heat of formations of ${\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)$ $$ $\mathrm{and}$ ${\mathrm{H}}_2\mathrm{O}$ $\left(\mathrm{l}\right)$ $$are -188 kJ/mol & -286 KJ/mol respectively. The change in the enthalpy of the reaction will be:

1.  – 196 kJ/mol

2.  + 196 kJ/mol

3.  + 948 kJ/mol

4.  – 948 kJ/mol