In the following process,
`H_(2)(g) to 2H(g), " "Delta_(H-H)H^(ө)=435.0" kJ "mol^(-1)` the enthalpy change involved is the

Options

A. bond dissociation enthalpy of H-H bond
B. bond association enthalpy of H-H bond
C. mean bond dissociation enthalpy of H-H bond
D. mean bond association enthalpy of H-H bond

Answer

A

Explanation

In the given equation,
`H_(2)(g) to 2H(g),Delta_(H-H)H^(ө)=435.0kJ" "mol^(-1)`
the enthalpy change involved is the bond dissociation enthalpy of H-H bond.