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 following reaction would be:

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

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

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)

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. \(-57.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

At standard conditions, if the change in the enthalpy for the following reaction is –109 kJ mol^–1 
H_2(g)+Br_2(g)$\to$2HBr_(g) and the bond energy of H₂ and Br₂ is 435 kJ mol^–1 and 192 kJ mol^–1 respectively, what is the bond energy (in kJ mol^–1) of HBr?

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:

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

The amount of heat needed to raise the temperature of 60.0 g of aluminium from 35°C to 55°C would be:

(Molar heat capacity of Al is \(24\) \(J\) \(\text{mol}^{- 1}\) \(K^{- 1}\))

The reaction of cyanamide, ${\mathrm{NH}}_2\mathrm{CN}$ $\left(\mathrm{s}\right)$ with dioxygen, was carried out in a bomb calorimeter, and ∆U was found to be $-742.7$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$ at 298 K:
\(\small{\mathrm{NH}_2 \mathrm{CN}(\mathrm{s})+\frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{g})+\mathrm{CO}_2(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})}\)

${}_{}$The enthalpy change for the reaction at 298 K would be:

$1.$ $-741.3$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$2.$ $+$ $753.9$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$3.$ $+$ $772.$ $7$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$4.$ $-845.$ $1$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$