The enthalpy of formation of all elements in their standard state is-

1.	Unity	2.	Zero
3.	Less than zero	4.	Different for each element

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 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:

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

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

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

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

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

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

If the volume of a gas is reduced to half from its original volume, then the specific heat will: