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

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

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

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  2A(g) + B(g) → 2D(g) ; ∆U° = - 10.5 kJ and  ∆S° = - 44.1 J K^-1, the value of ∆G°  for the given reaction would be-

1. 1.6 J

2. -0.16 kJ

3. 0.16 kJ

4. 1.6 kJ

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 reaction

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

$1.$ $326$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$2.$ $1304$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$3.$ $-328$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$4.$ $-1304$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

The standard enthalpy of the formation of CH₃OH_(l)  from the following data 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 $mo{l}^{-1}$ respectively.

The value of ${∆}_{\mathrm{r}}\mathrm{H}$ for the reaction would be-

${\mathrm{N}}_2{\mathrm{O}}_{4\left(\mathrm{g}\right)}+3{\mathrm{CO}}_{\left(\mathrm{g}\right)}\to {\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+3{\mathrm{CO}}_{2\left(\mathrm{g}\right)}$

$1.$ $-777.7$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$2.$ $$ $+777.7$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$3.$ $$ $+824.9$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$4.$ $-$ $345.4$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

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 $24$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$ ${\mathrm{K}}^{-1}$)

$1.$ $1.07$ $\mathrm{J}$
$2.$ $1.07$ $\mathrm{kJ}$
$3.$ $106.7$ $\mathrm{kJ}$
$4.$ $100.7$ $\mathrm{kJ}$

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 -

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