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

1.	741 KJ	2.	745 KJ
3.	720 KJ	4.	734 KJ

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

The incorrect expression among the following is :

\(\text { 1. In isothermal process, } \mathrm{W}_{\text {reversible }}=-\mathrm{nRT} \ln \frac{\mathrm{~V}_{\mathrm{f}}}{\mathrm{~V}_{\mathrm{i}}}\)
\(\text { 2. } \ln K=\frac{\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{~S}^{\circ}}{\mathrm{RT}}\)
\(\text { 3. } \mathrm{K}=\mathrm{e}^{-\Delta \mathrm{G}^{\circ} / \mathrm{RT}}\)
\(\text { 4. } \frac{\Delta \mathrm{G}_{\text {system }}}{\Delta \mathrm{S}_{\text {total }}}=-\mathrm{T}\)

For complete combustion of ethanol, C₂H₅OH_(l) + 3O_2(g) → 2 CO_2(g) + 3H₂O_(l) , the amount of heat produced as measured in a bomb calorimeter, is 1364.47 kJ mol^–1 at 25ºC. Assuming ideality the enthalpy of combustion, ∆_CH, for the reaction will be: (R = 8.314 J K^–1 mol^–1)

1.  – 1361.95 kJ mol–¹

2.  – 1460.50 kJ mol^–1

3.  – 1350.50 kJ mol^–1

4.  – 1366.95 kJ mol^–1

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 

Which of the following conditions is correct for a reversible reaction having both enthalpy change (ΔH, enthalpy change) and entropy change (ΔS, entropy change) 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 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.

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