The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are, –890.3 kJ mol^–1 , –393.5 kJ mol^–1, and –285.8 kJ mol^–1 respectively. The enthalpy of formation of CH₄(g) is-

 For the reaction A + B → C + D + q (kJ/mol), entropy change is positive. The reaction will be

1. Possible only at high temperature

2. Possible only at low temperature

3. Not possible at any temperature

4. Possible at any temperature

The enthalpy of combustion of carbon to CO₂ is –393.5 kJ mol^-1. The amount of heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas would be:

For the reaction, $2C{}^{}\to C{}^{}$, the value of ∆H and ∆S  are respectively:

1. $∆\mathrm{H}=0$ $,$ $∆\mathrm{S}=-\mathrm{ve}$

2. $∆\mathrm{H}=0$ $,$ $∆\mathrm{S}=0$

3. $∆\mathrm{H}=$ $-\mathrm{ve}$ $,$ $∆\mathrm{S}$ $=$ $-\mathrm{ve}$

4. $∆\mathrm{H}=$ $+\mathrm{ve}$ $,$ $∆\mathrm{S}=$ $+\mathrm{ve}$

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 \(24\) \(J\) \(\text{mol}^{- 1}\) \(K^{- 1}\))