question_answer 1)

The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, \[C{{O}_{2}}\](g) and\[{{H}_{2}}O(l)\] are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation,\[{{\Delta }_{f}}{{H}^{{}^\circ }}\]of benzene. Standard enthalpies of formation of \[C{{O}_{2}}\](g) and \[{{H}_{2}}O(l)\]are \[-393.5kJ\,mo{{l}^{-1}}\] and\[-285.83kJ\,mo{{l}^{-1}}\] respectively.

Answer:

Information shadow: (i) \[{{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g)\to 6C{{O}_{2}}(g)+3{{H}_{2}}O(l);\] \[\Delta H=-3267kJ\] (ii) \[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g);\,\,\,\,\Delta H=-393.5kJ\] (iii) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(l)\to {{H}_{2}}O(l);\,\,\,\,\,\Delta H=-285.83kJ\] The required equation is: \[6C(s)+3{{H}_{2}}(g)\to {{C}_{6}}{{H}_{6}}(l);\,\,\,\,\,\,\,\,\Delta H=?\] Problem solving strategy: The heat of required equation can be obtained by algebraic method. (ii) \[\times \] 6 + (iii) \[\times \] 3 + (i) Working it out: \[6C(s)+6{{O}_{2}}(g)\to 6C{{O}_{2}}(g);\Delta H=-393.5\times 6kJ\]\[3{{H}_{2}}(g)+\frac{3}{2}{{O}_{2}}(g)\to 3{{H}_{2}}(l);\,\,\,\,\Delta H=-285.83\times 3kJ\]\[6C{{O}_{2}}(g)+3{{H}_{2}}O(l)\to {{C}_{6}}{{H}_{6}}(l)+\frac{15}{2}{{O}_{2}}(g);\,\] \[\Delta H=+3267kJ\] On adding,_____________________________ \[6C(s)+3{{H}_{2}}(g)\to {{C}_{6}}{{H}_{6}}(l);\,\,\,\,\,\,\Delta H=+48.51kJ\,mo{{l}^{-1}}\]Alternatively, \[6C(s)+3{{H}_{2}}(g)\to {{C}_{6}}{{H}_{6}}(l)\] \[{{\Delta }_{r}}H=\sum \]Heat of combustion of reactants \[-\sum \] Heat of combustion of products = 6 x (-393.5) + 3 (-285.83) - (- 3267) \[=48.51kJ\,mo{{l}^{-1}}\]