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

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

The enthalpy changes for the following processes are listed below : 

Cl₂ (g) → 2Cl(g), 242.3 kJ mol^-1 

I₂ (g) → 2I (g), 151.0 kJ mol^-1 

ICI (g) → I(g) + Cl(g), 211.3 kJ mol^-1 

I₂ (s) → I₂ (g), 62.76 kJ mol^-1

Given that the standard states for iodine and chlorine  are I₂ (s) and Cl₂ (g), the standard enthalpy of formation of ICI (g) is : 

1. -14.6 kJ mol^-1 

2. -20.8 kJ mol^-1 

3. +16.8  kJ mol^-1 

4. +244.8  kJ mol^-1

Which of the following is the standard enthalpy of formation of carbon monoxide (CO), if the standard enthalpies of combustion of carbon (C) and carbon monoxide (CO) respectively are −393.5 kJ mol⁻¹  and −283 kJ mol⁻¹?

1. 110.5 kJ

2. 676.5 kJ

3. -676.5 kJ

4. -110.5 kJ