Which of the following is not correct?

1. $∆\mathrm{G}$ is zero for a reversible reaction.

2. $∆\mathrm{G}$ is positive for a spontaneous reaction.

3. $∆\mathrm{G}$ is negative for a spontaneous reaction.

4. $∆\mathrm{G}$ is positive for a non-spontaneous reaction.

A necessary condition for an adiabatic change is:

1. ∆T = 0

2. ∆P = 0

3. q = 0

4. w = 0

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

$∆\mathrm{U}°$ $$for combustion of methane is –x kJ mol^–1.

The value of $∆\mathrm{H}°$ for the same reaction would be:

$1.$ $=$ $∆\mathrm{U}°$
$2.$ $>$ $∆\mathrm{U}°$
$3.$ $<$ $∆\mathrm{U}°$
$4.$ $=0$

701 J of heat is absorbed by a system and 394 J of work is done by the system. The change in internal energy for the process is:

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:

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

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 reaction would be:

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

 ${\mathrm{N}}_2 + 3 {\mathrm{H}}_2 \to 2{\mathrm{NH}}_{3 ;} {∆}_{\mathrm{r}}{\mathrm{H}}^{°} = -92.4 \mathrm{kJ} {\mathrm{mol}}^{-1}$. 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: