If for a certain reaction ${∆}_{r}H$ is 30 kJ mol^–1 at 450 K, the value of ${∆}_{r}S$ (in JK^–1 mol^–1) for which the same reaction will be spontaneous at the same temperature is:

1.	70	2.	–33
3.	33	4.	–70

Consider the following processes:

                                       ∆H (kJ/mol)
½ A → B                            + 150
3B → 2C + D                      –125
E + A → 2D                        +350

For B + D → E + 2C, ∆H will be-

1. 325 kJ/mol 

2. 525 kJ/mol

3. –175 kJ.mol 

4. –325 kJ/mol

For vaporization of water at 1 atmospheric pressure, the values of ∆H and ∆S are 40.63 kJ mol^–1 and 108.8 JK^–1 mol^–1, respectively. The temperature when Gibbs energy change (∆G) for this transformation will be zero, is:

1. 393.4 K

2. 373.4 K

3. 293.4 K

4. 273.4 K

Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be:

1. 3 Joules

2. 9 Joules

3. Zero

4. Infinite

The following two reactions are known$F{e}_2{O}_3\left(s\right)+3CO\left(g\right)\to 2Fe\left(s\right)+3C{O}_2\left(g\right);$
$∆H=-26.88$ $kJ$
$FeO\left(s\right)+CO\left(g\right)\to Fe\left(s\right)+C{O}_2\left(g\right);$
$∆H=-16.5$ $kJ$

The value of $∆$H for the following reaction

$F{e}_2{O}_3\left(s\right)+CO\left(g\right)\to 2FeO\left(s\right)+C{O}_2\left(g\right)$ is:

1. -43.3 kJ

2. -10.3 kJ

3. +6.2 kJ

4. +10.3 kJ

Match List – I (Equations) with List – II (Type of processes) and select the correct option.

Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats (Cp/Cv) of the mixture will be:
1.  1.50
2.  3.3
3.  1.67
4.  0.83

Entropy change in surroundings when 1.00 mol of $H_2{O}_{\left(l\right)}$ is formed under standard conditions. ${∆}_f{H}^{\theta }=-286<mspace\; width="1em"></mspace>kJ<mspace\; width="1em"></mspace>mo{l}^{-1}$ is

$1.<mspace\; width="1em"></mspace>-959.73<mspace\; width="1em"></mspace>\mathrm{J}<mspace\; width="1em"></mspace>{\mathrm{K}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$
$2.<mspace\; width="1em"></mspace>234.98<mspace\; width="1em"></mspace>\mathrm{J}<mspace\; width="1em"></mspace>{\mathrm{K}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$
$3.<mspace\; width="1em"></mspace>959.73<mspace\; width="1em"></mspace>\mathrm{J}<mspace\; width="1em"></mspace>{\mathrm{K}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$
$4.<mspace\; width="1em"></mspace>312.56<mspace\; width="1em"></mspace>\mathrm{J}<mspace\; width="1em"></mspace>{\mathrm{K}}^{-1<mspace\; width="1em"></mspace>}{\mathrm{mol}}^{-1}$

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

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

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

The value of ∆G°  for the given reaction would be:
\( 2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow 2 \mathrm{D}(\mathrm{~g})\)
(Given: ∆U° = – 10.5 kJ and  ∆S° = – 44.1 J K^–1)