The standard enthalpies of combustion of C6H6, C(graphite) and H2(g) are respectively -3270 kJ mol-1 , -394 kJ mol-1 and -286 kJ mol-1. What is the standard enthalpy of formation of C6 H6(l) in kJ mol-1 ?

Question

The standard enthalpies of combustion of C6H6, C(graphite) and H2(g) are respectively -3270 kJ mol-1 , -394 kJ mol-1 and -286 kJ mol-1. What is the standard enthalpy of formation of C6 H6(l) in kJ mol-1 ? (A) - 48 (B) + 48 (C) - 480 (D) + 480

Tardigrade Admin · Accepted Answer

On inverting Eq. (i), we get 6 CO 2+3 H 2 O longrightarrow C 6 H 6+(15/2) O 2 Δ H=+3270 kJ mol -1 Addition of Eq. (iv), (v) and (vi) gives 6 C ( s )+3 H 2 longrightarrow C 6 H 6 Δ H =+3270+(-2364-858) =+48 kJ mol -1 Thus, the standard enthalpy of formation of C 6 H 6,(Δf H C 6 H 6) is +48 kJ mol -1

Q. The standard enthalpies of combustion of $C_{6} H_{6}, C_{(g r a p hi t e)}$ and $H_{2} (g)$ are respectively $- 3270 k J$ $m o l^{- 1}$ , $- 394 k J m o l^{- 1}$ and $- 286 k J m o l^{- 1}$ . What is the standard enthalpy of formation of $C_{6} H_{6} (l)$ in $k J m o l^{- 1}$ ?

$- 48$

$+ 48$

$- 480$

$+ 480$

$- 72$

Q. The standard enthalpies of combustion of C6​H6​,C(graphite)​ and H2​(g) are respectively −3270kJ mol−1 , −394kJmol−1 and −286kJmol−1. What is the standard enthalpy of formation of C6​H6​(l) in kJmol−1 ?

−48

+48

−480

+480

−72