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The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO_(2(g)) and H_(2) O_((l)) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, Delta_(r) H^( Theta ) of benzene. Standard enthalpies of formation of CO_(2(g)) and H_(2) O_((l)) are -393.5 kJ "mol"^(-1) and -285.83 "kJ mol"^(-1) respectively. |
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Answer» Solution :The formation reaction of benzene is given by `6C_("(graphite)") + 3H_(2(g)) to C_(6) H_(6(l)), Delta_(f) H^( Theta )= (?)….(i)` Then enthalpy of combustion of 1 MOL of benzene is : `C_(6) H_(6(l)) + (15)/(2) O_(2) to 6CO_(2(g)) + 3H_(2) O_((l)),` `Delta_(C) H^( Theta ) = - 3267 "kJ mol"^(-1) ...(ii)` The enthalpy of formation of 1 mol of `CO_(2(g))`, `C_("(graphite)") + O_(2(g)) to CO_(2(g)) ,` `Delta_(f) H^( Theta ) = -393.5 "kJ mol"^(-1) ...(III)` The enthalpy of formation of 1 mol of `H_(2) O_((l))` is `H_(2(g)) + (1)/(2) O_(2(g)) to H_(2) O_((l)),` `Delta_(f) H^( Theta ) = -285.83 "kJ mol"^(-1) ....(iv)` Multiplying eq. (iii) by 6 and eq. (iv) by 3 we GET: `6C_("(graphite)") + 6O_(2(g)) to 6CO_(2(g)),` `Delta_(f) H^( Theta ) =-2361 "kJ mol"^(-1)` `3H_(2(g)) + (3)/(2) O_(2(g)) to 2H_(2) O_((l)),` `Delta_(f) H^( Theta )= -857.49 "kJ mol"^(-1)` SUMMING up the above two equations : `6C_("(graphite)") + 3H_(2(g)) + (15)/(2) O_(2(g)) to 6CO_(2(g)) + 3H_(2)O_((l)),` `Delta_(f) H^( Theta ) = -3218.49 "kJ mol"^(-1) .....(V)` Reversing equation (ii), `6 CO_(2(g)) + 3H_(2) O_((l)) to C_(6) H_(6(l)) + (15)/(2) O_(2(g)),` `Delta_(f) H^( Theta ) = 3267.0 "kJ mol"^(-1) ....(vi)` Adding equations (v) and (vi), we get, `6C_("(graphite)") + 3H_(2(g)) to C_(6)H_(6(l)),` `Delta_(f) H^( Theta ) = 48.51 "kJ mol"^(-1)` |
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