If, for a dimerization reaction, 2A_(g) → A_2(g)  at   298 K , ∆U^Θ = -20 kJ mol^-1   ∆S^Θ   = - 30 J K^-1mol^-1 , then ∆G^Θ  will be:

1. -10. 4 kJ

2. 18.9 kJ

3. -13.5 kJ

4. 17. 4 kJ

Assuming ideal behaviour, the magnitude of log K for the following reaction at 25°C is
x × 10^–1 . The value of x is:

$3\mathrm{HC}\equiv {\mathrm{CH}}_{\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{C}}_6{{\mathrm{H}}_6}_{\left(\mathrm{l}\right)}$

$[\text{\hspace{0.17em}Given:\hspace{0.17em}}$
${\mathrm{\Delta }}_f{\mathrm{G}}^{\circ }(\mathrm{HC}\equiv \mathrm{CH})=-2.04\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1}$
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\circ }\left({\mathrm{C}}_6{\mathrm{H}}_6\right)=-1.24\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1}$

$\mathrm{R}=8.314\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}]$

1. 860

2. 875

3. 855

4. 895

The average S–F bond energy in kJ mol^–1 of SF₆ is:

 [The values of standard enthalpy of formation of
SF₆(g), S(g), and F(g) are –1100, 275, and 80 kJmol^–1 respectively.]

The reaction of cyanamide, NH₂CN_(s) with oxygen was run in a bomb calorimeter and $∆$U was found to be –742.24 kJ mol^–1. The magnitude of ${\mathrm{\Delta H}}_{298}$(KJ) for the given-below reaction is:

NH₂CN_(s) + \(\frac{3}{2}\)O_2(g) → N_2(g) + O_2(g) + H₂O_(l) 

[Assume ideal gases and \(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)]$$

Given
\(\begin{aligned} &\mathrm{{C}_{{(graphite) }}+{O}_{2}({~g})} → \mathrm{{CO}_{2}({~g})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=-393.5 {~kJ} {~mol}^{-1}} \\ &\mathrm{H_{2}(g) + \frac{1}{2} {O}_{2}({~g})} → \mathrm{{H}_{2} {O}({l})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=-285.8 {~kJ} {~mol}^{-1}} \\ &\mathrm{{CO}_{2}({~g})+2 {H}_{2} {O}({l})} → \mathrm{{CH}_{4}({~g})+2 {O}_{2}({~g})} \\ &\mathrm{\Delta_{r} {H}^{\circ}=+890.3 {~kJ} {~mol}^{-1}} \end{aligned}\)

Based on the above thermochemical equations, the value of Δ_rH° at 298 K for the reaction
\(\mathrm{C_{(graphite)} + 2 H_{2} (g) → CH_{4} (g)}\) will  be :

1. –74.8 kJ mol^–1

2. –144.0 kJ mol^–1

3. +74.8 kJ mol^–1

4. +144.0 kJ mol^–1

The incorrect expression among the following is -

1. In isothermal process, ${\mathrm{W}}_{\text{reversible\hspace{0.17em}}}=-\mathrm{nRT}<mspace\; width="1em"></mspace>\ln <mspace\; width="1em"></mspace>\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}$

2. $\ln K=\frac{{\mathrm{\Delta H}}^{\circ }-{\mathrm{T\Delta S}}^{\circ }}{\mathrm{RT}}$

3. $\mathrm{K}={\mathrm{e}}^{-{\mathrm{\Delta G}}^{\circ }/\mathrm{RT}}$

4. $\frac{{\mathrm{\Delta G}}_{\text{system\hspace{0.17em}}}}{{\mathrm{\Delta S}}_{\text{total\hspace{0.17em}}}}=-\mathrm{T}$

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm³ to a volume of 100 dm³ at 27º C is :

1. 38.3 J mol⁻¹ K⁻¹

2. 35.8 J mol⁻¹ K⁻¹

3. 32.3 J mol⁻¹ K⁻¹

4. 42.3  J mol⁻¹ K⁻¹ 

For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If T_e is the temperature at equilibrium, the reaction would be spontaneous when:

1.  T = T_e

2.  T_e > T

3.  T > T_e

4.  T_e is 5 times T

The standard enthalpy of formation of NH₃ is -46 0. kJ mol⁻¹. If the enthalpy of formation of H₂ from its atoms is -436 kJ mol⁻¹ and that of N₂ is -712 kJ mol^-1, the average bond enthalpy of N − H bond in NH₃ is:

1. $-1102\mathrm{kJ}{\mathrm{mol}}^{-1}$

2. $-964\mathrm{kJ}{\mathrm{mol}}^{-1}$

3. $+352\mathrm{kJ}{\mathrm{mol}}^{-1}$

4. $+1056\mathrm{kJ}{\mathrm{mol}}^{-1}$

The value of enthalpy change ($∆$H) for the reaction

${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{\left(1\right)}+3{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$ at $27°C$ is -1366.5 kJ mol^-1.