Explanation:
Reaction between chlorine and hydrogen gas will take place as follows.
[tex]Cl_{2}(g) + H_{2}(g) \rightarrow 2HCl(g)[/tex]
Since, standard enthalpy of [tex]Cl_{2}(g)[/tex] is 0 kJ/mol and [tex]H_{2}(g)[/tex] is also 0 kJ/mol. [tex]\Delta H_{HCl}[/tex] is -92.30 kJ/mol.
Hence, change in enthalpy will be calculated as follows.
[tex]\Delta H = \Delta H_{products} - \Delta H_{reactants}[/tex]
= (-92.30 - 0) kJ/mol
= -ve
Also, formation of bonds is taking place between both chlorine and hydrogen atoms. As a result, heat will be released during this process.
Thus, we can conclude that change in enthalpy will be negative.
