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

From the following bond energies:
H—H bond energy: 431.37 kJ mol^-1 
C=C bond energy: 606.10 kJ mol^-1 
C—C bond energy: 336.49 kJ mol^-1 
C—H bond energy: 410.50 kJ mol^-1 
Enthalpy for the reaction, 

will be:

$∆{\mathrm{H}}_{\mathrm{f}}^0$ (298K) of methanol is given by the chemical equation -

1. C(diamond)+$\frac{1}{2}{O}_2$+2H₂$\to$CH₃OH

2. CH₄+$\frac{1}{2}{O}_2$$\to$CH₃OH

3. CO+2H₂$\to$CH₃OH

4. C(graphite)+$\frac{1}{2}{O}_2$+2H₂$\to$CH₃OH

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-

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

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

Which of the following is not an endothermic reaction?

1. Combustion of methane

2. Decomposition of water

3. Dehydrogenation of ethane or ethylene

4. Conversion of graphite to diamond

When 4 g of iron is burnt to ferric oxide at a constant pressure, 29.28 kJ of heat is evolved.

The enthalpy of formation of ferric oxide will be-

(At. mass of Fe = 56) ?

1. $-$81.98 kJ

2. $-$ 819.8 kJ

3. $-$ 40.99 kJ

4.  +819.8 kJ

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