Consider the following diagram for a reaction $\mathrm{A}\to \mathrm{C}$. 

The nature of the reaction is-

1. Exothermic

2. Endothermic

3. Reaction at equilibrium

4. None of the above

Assume each reaction is carried out in an open container. For which of the following reactions will 
∆ H be equal to ∆ U ?

1. ${\mathrm{PCl}}_3\left(\mathrm{g}\right)$ $\to {\mathrm{PCl}}_3\left(\mathrm{g}\right)$ $+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$ $$

2. $2\mathrm{CO}\left(\mathrm{g}\right)$ $+{\mathrm{O}}_2\left(\mathrm{g}\right)$ $\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)$ $$

3. ${\mathrm{H}}_2\left(\mathrm{g}\right)$ $+{\mathrm{Br}}_2\left(\mathrm{g}\right)$ $\to 2\mathrm{HBr}\left(\mathrm{g}\right)$

4. $\mathrm{C}\left(\mathrm{s}\right)$ $+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$ $\to 2{\mathrm{H}}_2\left(\mathrm{g}\right)$ $+{\mathrm{CO}}_2\left(\mathrm{g}\right)\hspace{0.17em}$

For the reaction, $2{\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{N}}_2\mathrm{O}\left(\mathrm{g}\right)$, at 298K $∆\mathrm{H}$ is 164 kJ mol^-1. The $∆\mathrm{E}$ of the reaction is :
1. \(166.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
2. \(141.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
3. \(104.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
4. \(-169 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The bond energies of $\mathrm{C}\equiv \mathrm{C}$, C-H, H-H, and C=C are 198, 98, 103, and 145 kcal respectively.

The enthalpy change of the reaction $\mathrm{HC}\equiv \mathrm{CH}+{\mathrm{H}}_2\to {\mathrm{C}}_2{\mathrm{H}}_4$ would be:

1. 48 kcal

2. 96 kcal

3. -40 kcal

4. -152 kcal

For the following given equations and $∆\mathrm{H}°$ values, determine the enthalpy of reaction at 298 K for the reaction:

C₂H₄(g) + 6F₂(g) $\to$ 2CF₄(g) + 4HF(g)

H₂(g) + F₂(g) $\to$ 2HF(g)       $∆{\mathrm{H}}_1^{°}$= -537 kJ

C(s) + 2F₂(g) $\to$CF₄(g)         $∆{\mathrm{H}}_2^{°}$=-680 kJ

2C(s) + 2H₂(g) $\to$C₂H₄(g)    $∆{\mathrm{H}}_3^{°}$= 52 kJ

1. –1165 kJ

2. –2486 kJ

3. +1165 kJ

4. +2486 kJ