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

An enthalpy diagram for a particular reaction is given below:

The correct statement among the following is-

1. Reaction is spontaneous

2. Reaction is non-spontaneous

3. Cannot predict spontaneity of the reaction from the graph given above

4. None of the above

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

Consider the following graph.

The work done shown by the above-mentioned graph is-

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

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

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

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

The thermodynamic stability of NO_(g) based on the above data is:

1. Less than NO_2(g) 

2. More than NO_2(g)

3. Equal to NO_2(g)

4. Insufficient data

For the reaction:

\(C_{3} H_{8} \left(\right. g \left.\right) + 5 O_{2} \left(\right. g \left.\right) \rightarrow 3 CO_{2} \left(\right. g \left.\right) + 4 H_{2} O \left(\right. l \left.\right)\) at constant temperature, ∆H – ∆E is:

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