The standard enthalpies of combustion of `C_(6)H_(6)(l),C_(("graphite"))andH_(2)(g)` are respectively `-3270kJmol^(-1),-394kJmol^(-1) and -286kJmol^(-1)`. What is the standard enthalpy of formation of `C_(6)H_(6)(l)` in kJ `mol^(-1)`?

Options

A. `-48`
B. `+48`
C. `-480`
D. `+480`

Answer

B

Explanation

Given,
(i) `C_(6)H_(6)+(15)/(2)O_(2)rarr6CO_(2)+3H_(2)O, DeltaH=-3270kJmol^(-1)`
(ii) C(graphite) `+O_(2)rarrCO_(2), DeltaH=-394kJmol^(-1)`
(iii) `H_(2)+(1)/(2)O_(2)rarrH_(2)O, DeltaH=-286kJmol^(-1)`
Multiplication of Eq. (ii) by 6 and Eq. (iii) by 3, gives
(iv) `6C(s)+6O_(2)rarr6CO_(2), DeltaH=-394xx6kJmol^(-1)`
`=-2364 kJmol^(-1)`
(iv) `3H_(2)+(3)/(2)O_(2)rarr3H_(2)O, DeltaH=-286xx3kJmol^(-1)`
`=-858kJmol^(-1)`
On inverting Eq. (i), we get
(vi) `6CO_(2)+3H_(2)OrarrC_(6)H_(6)+(15)/(2)O_(2), DeltaH=+3270kJmol^(-1)`
Addition of Eq. (iv), (v) and (vi) gives
`6C(s)+3H_(2)rarrC_(6)H_(6), DeltaH=+3270+(-2364-858)`
`=+48kJmol^(-1)`
Thus, the standard enthalpy of formation of `C_(6)H_(6),(Delta_(f)H_(C_(6)H_(6)))` is `+48kJmol^(-1)`.