The standard enthalpy of formation of `H_(2)(g)` and `Cl_(2)(g)` and `HCl(g)` are `218 kJ//mol`, `121.88 kJ//mol` and `-93.31 kJ//mol` respectively. Calculate standard enthalpy change in `kJ` for
`(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g)`

Options

A. `+ 431.99`
B. `-262.14`
C. `-431.99`
D. `+247.37`

Answer

B

Explanation

`Delta_(r) H= Delta H_("products")- Delta H_("reactants") = -92.3 - [(218)/(2) + (121.68)/(2)] = -92.3 - [109 + 60.84]`
`= -262.14 kJ`/mol