`Delta_(f) H` of graphite is `0.23 kJ mol^(-1)` and `Delta_(f) H` of diamond is `1.896 kh mol^(-1)`. `Delta H_(transition)` from graphite to diamond is

Options

A. `1.66 kJ mol^(-1)`
B. `2.1 kJ mol^(-1)`
C. `2.33 kJ mol^(-1)`
D. `1.5 kJ mol^(-1)`

Answer

A

Explanation

`C` (graphite) `rarr C` (diamond)
`Delta_(r) H^(@)` (diamond) - `Delta_(f) H^(@)` (products) `- sum b_(i) Delta_(f) H^^(@)` (reactants)
`Delta_(r) H^(@)` (diamond) - `Delta_(f) H^(@)` (graphite)
`(1.896) - 0.23)`
`= 1.666 kJ mol^(-1)`