The heat of combustion of ethanol into carbon dioxides and water is –327 kcal at constant pressure. The heat evolved (in cal) at constant volume and 27°C (if all gases behave ideally) is:
(R = 2 cal mol^–1 K^–1)
1. 316 kcal
2. 326 kcal
3. 365 kcal
4. 342 kcal

For one mole of an ideal gas, which of these statements must be true?
(I) U and H each depend only on temperature.
(II) Compressibility factor z is not equal to 1.
(III) C_{P, m} – C_{V, m} = R
(IV) dU = C_VdT for any process.
1. (I), (III) and (IV)
2. (II), (III) and (IV)
3. (III) and (IV)
4. (I) and (III)

Five moles of an ideal gas at 1 bar and 298 K are expanded into a vacuum till the volume doubles. The work done is:
1. –RT ln V₂/V₁
2. C_V(T₂ – T₁)
3. zero
4. – RT(V₂ – V₁)

Lattice energy and enthalpy of the solution of NaCl are 788 kJ mol^–1 and 4 kJ mol^–1 , respectively. The hydration enthalpy of NaCl is:
1. –780 kJ mol^–1
2. –784 kJ mol^–1
3. 780 kJ mol^–1
4.  784 kJ mol^–1

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

ΔU is equal to:

1. Adiabatic work

2. Isothermal work

3. Isochoric work

4. Isobaric work