The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O(l) are produced and 3267 kJ of heat is liberated. Standard enthalpies of formation of CO2(g) and H2O(l) are -393.5 kJ mol-1 and -285.83 kJ mol-1 respectively. The standard enthalpy of formation, Δf H° of benzene is

Question

The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O(l) are produced and 3267 kJ of heat is liberated. Standard enthalpies of formation of CO2(g) and H2O(l) are -393.5 kJ mol-1 and -285.83 kJ mol-1 respectively. The standard enthalpy of formation, Δf H° of benzene is (A) 45.5 kJ mol-1 (B) 20 kJ mol-1 (C) 48.51 kJ mol-1 (D) 15.52 kJ mol-1

Tardigrade Admin · Accepted Answer

Required equation is 6C(s)+3H2(g) arrow C6H6(l) ; Δ Hf = ? Given: (i) C6H6(l)+(15/2)O2(g)arrow6CO2(g)+3H2O(l), Δ H=-3267 kJ mol-1 (ii) C(s)+O2(g)arrow, CO2(g), Δ H =-393.5 kJ mol-1 (iii) H2(g)+(1/2)O2(g) arrow H2O(l), Δ H=-285.83 kJ mol-1 Operating 6 × eqn. (ii) +3× eqn. (iii)-eqn. (i): 6C(s)+3H2(g)arrow C6H6(l), Δ H = 6(-395.5)+3(-285.83)-(-3267.0) =-2361-857.49+3267.0 =48.51 kJ mol-1

$45.5 k J m o l^{- 1}$

$20 k J m o l^{- 1}$

$48.51 k J m o l^{- 1}$

$15.52 k J m o l^{- 1}$

45.5kJmol−1

20kJmol−1

48.51kJmol−1

15.52kJmol−1

$45.5 k J m o l^{- 1}$

$20 k J m o l^{- 1}$

$48.51 k J m o l^{- 1}$

$15.52 k J m o l^{- 1}$