The standard enthalpy of vaporization ${∆}_{\mathrm{vap}}{\mathrm{H}}^{\mathrm{o}}$ for water at 100 ^oC is 40.66 kJ mol^–1. The internal energy of vaporization of water at 100 ^oC (in kJ mol^–1) is: 
(Assume water vapour behaves like an ideal gas.)

Given the following reaction:
\(4H(g)\)→  \(2 H_{2}\)\((g)\)
The enthalpy change for the reaction is -869.6 kJ. The dissociation energy of the H-H bond is:
1. -869.6 kJ
2. +434.8kJ
3. +217.4kJ
4. -434.8 kJ

Consider the following processes:

                                       ∆H (kJ/mol)
½ A → B                            + 150
3B → 2C + D                      –125
E + A → 2D                        +350

For B + D → E + 2C, ∆H will be-

1. 325 kJ/mol 

2. 525 kJ/mol

3. –175 kJ.mol 

4. –325 kJ/mol

The following two reactions are known$F{e}_2{O}_3\left(s\right)+3CO\left(g\right)\to 2Fe\left(s\right)+3C{O}_2\left(g\right);$
$∆H=-26.88$ $kJ$
$FeO\left(s\right)+CO\left(g\right)\to Fe\left(s\right)+C{O}_2\left(g\right);$
$∆H=-16.5$ $kJ$

The value of $∆$H for the following reaction

$F{e}_2{O}_3\left(s\right)+CO\left(g\right)\to 2FeO\left(s\right)+C{O}_2\left(g\right)$ is:

1. -43.3 kJ

2. -10.3 kJ

3. +6.2 kJ

4. +10.3 kJ

The difference between enthalpy (∆H) and internal energy (∆E) for the given below reaction,
under constant temperature, is:
\(C_{3} H_{8} \left(\right. g \left.\right) + 5 O_{2} \left(\right. g \left.\right) \rightarrow 3 CO_{2} \left(\right. g \left.\right) + 4 H_{2} O \left(\right. l \left.\right)\)

For the given reaction 

$2H_2{O}_2\left(l\right)$ $\to$ $2H_{2}O\left(l\right)$ $+$ $O_2\left(g\right)$, the heat of formations of ${\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)$ $$ $\mathrm{and}$ ${\mathrm{H}}_2\mathrm{O}$ $\left(\mathrm{l}\right)$ $$are -188 kJ/mol & -286 KJ/mol respectively. The change in the enthalpy of the reaction will be:

1.  – 196 kJ/mol

2.  + 196 kJ/mol

3.  + 948 kJ/mol

4.  – 948 kJ/mol