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

\(\small{\mathrm{CH}_3 \mathrm{OH}_{(l)}+\frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}_{(l)} \text {; }}\) \( \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-726 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}\)

\(\small{\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \text {; } }\) \(\Delta_{\mathrm{c}} \mathrm{H}^{\circ}=-393 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}\)

\(\small{\mathrm{H}_{2(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{H}_2 \mathrm{O}_{(l)} \text {; } } \) \(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

 

1.	−239 kJ mol−1	2.	+239 kJ mol−1
3.	−47 kJ mol−1	4.	+47 kJ mol−1

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

The enthalpy change for the reaction

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

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

For the reaction at 298 K,

2A + B → C

ΔH = 400 kJ mol⁻¹ and ΔS = 0.2 kJ K⁻¹ mol⁻¹. The reaction will become spontaneous at:

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)

The equilibrium constant for a reaction is 10. The value of $∆\mathrm{G}°$ will be:

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

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

Entropy change in surroundings when 1.00 mol of $H_2{O}_{\left(l\right)}$ is formed under standard conditions. ${∆}_f{H}^{\theta }=-286 kJ mo{l}^{-1}$ is

$1. -959.73 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$
$2. 234.98 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$
$3. 959.73 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$
$4. 312.56 \mathrm{J} {\mathrm{K}}^{-1 }{\mathrm{mol}}^{-1}$

If for a certain reaction ${∆}_{r}H$ is 30 kJ mol^–1 at 450 K, the value of ${∆}_{r}S$ (in JK^–1 mol^–1) for which the same reaction will be spontaneous at the same temperature is:

 ${\mathrm{N}}_2 + 3 {\mathrm{H}}_2 \to 2{\mathrm{NH}}_{3 ;} {∆}_{\mathrm{r}}{\mathrm{H}}^{°} = -92.4 \mathrm{kJ} {\mathrm{mol}}^{-1}$. The standard enthalpy of formation of ${\mathrm{NH}}_3$ gas in the above reaction would be:

The enthalpy of formation of ${\mathrm{CO}}_{\left(\mathrm{g}\right)},$ ${\mathrm{CO}}_{2\left(\mathrm{g}\right)},$ ${\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)} , \mathrm{and}$ ${\mathrm{N}}_2{\mathrm{O}}_{4\left(\mathrm{g}\right)}$ $$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:

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