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

Consider the given reaction:
\(2 \mathrm{Cl}(\mathrm{~g}) \rightarrow \mathrm{Cl}_2(\mathrm{~g})\)

What are the values of \(∆\mathrm{H}\) and \(∆\mathrm{S}\), respectively?
1. \(\Delta \mathrm{H}=0, \Delta \mathrm{~S}=-\mathrm{ve}\)
2. \(\Delta \mathrm{H}=0, \Delta \mathrm{~S}=0\)
3. \(\Delta \mathrm{H}=-\mathrm{ve}, \Delta \mathrm{~S}=-\mathrm{ve}\)
4. \(\Delta \mathrm{H}=+\mathrm{ve}, \Delta \mathrm{~S}=+\mathrm{ve}\)

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-

The correct statement among the following is:

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

${∆}_{\mathrm{f}}{\mathrm{U}}^{⊝}$ of formation of ${\mathrm{CH}}_4\left(\mathrm{g}\right)$ at certain temperature is $-393$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$. The value of ${∆}_{\mathrm{f}}{\mathrm{H}}^{⊝}$ is-

1. Zero

2. $<{∆}_{\mathrm{f}}{\mathrm{U}}^{⊝}$

3. $>{∆}_{\mathrm{f}}{\mathrm{U}}^{⊝}$

4. Equal to ${∆}_{\mathrm{f}}{\mathrm{U}}^{⊝}$

In an adiabatic process, no transfer of heat takes place between the system and its surroundings. The correct option for free expansion of an ideal gas under adiabatic condition from the following is:

1. $\mathrm{q}=0,<mspace\; width="1em"></mspace>∆\mathrm{T}\ne 0,<mspace\; width="1em"></mspace>\mathrm{W}=0$

2. $\mathrm{q}\ne 0,<mspace\; width="1em"></mspace>∆\mathrm{T}=0,<mspace\; width="1em"></mspace>\mathrm{W}=0$

3. $\mathrm{q}=0,<mspace\; width="1em"></mspace>∆\mathrm{T}=0,<mspace\; width="1em"></mspace>\mathrm{W}=0$

4. $\mathrm{q}=0,<mspace\; width="1em"></mspace>∆\mathrm{T}=0,<mspace\; width="1em"></mspace>\mathrm{W}\ne 0$

The pressure-volume work for an ideal gas can be calculated by using the expression $\mathrm{W}={\int }_{{\mathrm{V}}_{\mathrm{i}}}^{{\mathrm{V}}_{\mathrm{f}}}{\mathrm{p}}_{\mathrm{ex}}d\mathrm{V}$.

The work can also be calculated from the pV-plot by using the area under the curve within the specified limits.
An ideal gas is compressed (a) reversibly or (b) irreversibly from volume ${\mathrm{V}}_{\mathrm{i}}$ to ${\mathrm{V}}_{\mathrm{f}}$.
The correct option is:

1. $\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{reversible}\right)=\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{irreversible}\right)$

2. $\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{reversible}\right)<\mathrm{W}\hspace{0.17em}\left(\mathrm{irreversible}\right)$

3. $\mathrm{W}<mspace\; width="1em"></mspace>(\mathrm{reversible})>\mathrm{W}<mspace\; width="1em"></mspace>(\mathrm{irreversible})$

4. $\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{reversible}\right)=\mathrm{W}<mspace\; width="1em"></mspace>\left(\mathrm{irreversible}\right)+{\mathrm{p}}_{\mathrm{ex}}.∆\mathrm{V}$

The entropy change can be calculated by using the expression $∆\mathrm{S}=\frac{{\mathrm{q}}_{\mathrm{rev}}}{\mathrm{T}}$. When water freezes in a glass beaker, the correct statement among the following is:

Based on the reactions, ascertain the valid algebraic relationship:
$\left(i\right)$ $C\left(g\right)$ $+4H$ $\left(g\right)$ $\to C{H}_4\left(g\right);$ $$
$∆H_r$ $=x$ $kJ$ $mo{l}^{-1}$
$\left(ii\right)$ $C\left(graphite\right)$ $+$ $2H_2$ $\left(g\right)$ $\to$ $$ $C{H}_4;$ $$
$∆H_r$ $$ $=$ $y$ $kJ$ $mo{l}^{-1}$