question_answer 1)

  During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermo chemical reaction for above change is : (a) \[2{{C}_{4}}{{H}_{10}}(g)+13{{O}_{2}}(g)\to 8C{{O}_{2}}(g)+10{{H}_{2}}O(l)\]\[{{\Delta }_{c}}H=-2658.0kJ\,mo{{l}^{-1}}\] (b) \[{{C}_{4}}{{H}_{10}}(g)+\frac{13}{2}{{O}_{2}}(g)\to 4C{{O}_{2}}(g)+5{{H}_{2}}O(g)\]                \[{{\Delta }_{c}}H=-1329.0kJ\,mo{{l}^{-1}}\] (c) \[{{C}_{4}}{{H}_{10}}(g)+\frac{13}{2}{{O}_{2}}(g)\to 4C{{O}_{2}}(g)+5{{H}_{2}}O(l)\]  \[{{\Delta }_{c}}H=-2658.0kJ\,mo{{l}^{-1}}\] (d) \[{{C}_{4}}{{H}_{10}}(g)+\frac{13}{2}{{O}_{2}}(g)\to 4C{{O}_{2}}(g)+5{{H}_{2}}O(l)\] \[{{\Delta }_{c}}H=+2658.0\,kJ\,mo{{l}^{-1}}\]

Answer:

  (c) Exothermic reaction for combustion of one mole of butane is represented as : \[{{C}_{4}}{{H}_{10}}(g)+\frac{13}{2}{{O}_{2}}(g)\to 4C{{O}_{2}}(g)+5{{H}_{2}}O(l)\] \[{{\Delta }_{c}}H=-2658kJ\,mo{{l}^{-1}}\]