The heat of atomisation of `PH_(3(g))` is `228kcal mol^(-1)` and that of `P_(2)H_(4)` is `355kcal mol^(-1)`. Calculate the average bond energy of `P-P` bond.

Options

A. 63 kcal/mol
B. 58 kcal/mol
C. 5 kcal/mol
D. 45 kcal/mol

Answer

C

Explanation

`PH_(3(g)) rarr P_((g)) + 3H_((g)) Delta H_1 = 228 kcal//mol`
`H_2 P - PH_2 rarr 2P + 4H_((g)) Delta H_2 = 355 kcal//mol`
`Delta H_2 = E_(p-p) + 4 E_(P-H)`
`355 = E_(p-p) + 4 xx (228)/(3)`
`E_(P-P) = 51 kcal//mol`.