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.

The value of internal energy change for the above reaction at this temperature will be:
1. –1369.0 k
2. –1364.0 kJ
3. –1361.5 k
4. 1371.5 kJ

Consider the reaction:

$\mathrm{4NO2(g) \; + \; O2(g) \; \to \; 2N2O5(g)}$ Δ_rH = –111 kJ

If N₂O_5(s) is formed instead of N₂O_5(g) in the above reaction, the $∆$_rH value will be :
(Given: $∆$H of sublimation for N₂O₅ is 54 kJ mol^–1)

1. +54 kJ

2. + 219 kJ

3. –219 kJ

4. –165 kJ

Find the standard enthalpy of formation of OH⁻(aq) at 25°C from the following data:

ΔfH°[H⁺(aq)] = 0

H₂O(l) → H⁺(aq) + OH⁻(aq)  ΔH° = +57.32 kJ mol⁻¹

H₂(g) + ½O₂(g) → H₂O(l)  ΔH° = –286.20 kJ mol⁻¹
 

1. –22.88 kJ

2. –228.88 kJ

3. +228.88 kJ

4. –343.52 kJ

Standard entropy of ${\mathrm{X}}_2,{\mathrm{Y}}_2$ and ${\mathrm{XY}}_3$ are $60,40$ and $50\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1},$ respectively. For the reaction, $\frac{1}{2}<mspace\; width="1em"></mspace>{\mathrm{X}}_2+\frac{3}{2}<mspace\; width="1em"></mspace>{\mathrm{Y}}_2<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>{\mathrm{XY}}_3∆\mathrm{H}=-30<mspace\; width="1em"></mspace>\mathrm{kJ},$ to be at equilibrium, the temperature will be : 

1. 500 K

2. 750 K

3. 1000 K

4. 1250 K

Assuming that water vapor is an ideal gas, the internal energy change (∆U) when 1 mol of water is vaporized at 1 bar pressure and 100°C, will be:

(Given: Molar enthalpy of vaporization of water at 1 bar and 373 K = 41 kJ mol^–1 and R = 8.3 J mol^–1 K^–1) 

1. 4.100 kJ mol^–1

2. 3.7904 kJ mol^–1

3. 37.904 kJ mol^–1

4. 41.00 kJ mol^–1

Consider the following reaction for the conversion of limestone to lime:

CaCO₃​(s)→CaO(s)+CO₂​(g)

The values of ∆Hº and ∆Sº are +179.1 kJ/mol and 160.2 J/K, respectively, at 298 K and 1 bar. Assuming that ∆Hº and ∆Sº remain constant with temperature, calculate the temperature above which the conversion of limestone to lime will be spontaneous.

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

Which of the following conditions is correct for a reversible reaction to be spontaneous, having both enthalpy change (ΔH) and entropy change (ΔS) positive, if Tₑ is the equilibrium temperature?

1. T = Te (reaction is at equilibrium condition)

2.  Te > T (equilibrium temperature is greater than reaction temperature)

3.  T > Te (reaction temperature is greater than equilibrium temperature)

4.  Te = 5T (equilibrium temperature is five times reaction temperature)

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