A) - 6.2kJ
B) - 37.4kJ
C) - 35.5kJ
D) - 20.0kJ
Correct Answer: B
Solution :
[b] ? 37.4kJ \[{{C}_{8}}{{H}_{18}}+\frac{25}{2}{{O}_{2}}\to 8C{{O}_{2}}+9{{H}_{2}}O\] \[\Delta H{}^\circ \,reaction\,=[8\times \Delta H_{f}^{o}(C{{O}_{2}})+9\Delta H_{f}^{o}({{H}_{2}}O)]\] \[-\left[ \Delta H_{f}^{o}({{C}_{8}}{{H}_{18}})+\frac{25}{2}\Delta H_{f}^{o}({{O}_{2}})] \right]\] \[=\left[ 8\,(-490)+9(-240) \right]-\left[ (160)+\frac{25}{2}Co \right]\] \[=-3920-2160-160=6420\,Jmo{{l}^{-1}}\] \[=-6240\,J/mo{{l}^{-1}}\] \[\therefore \] For 6 moles, enthalpy of combustion, \[=-6240\times 6\] \[=-37400J=-3704KJ\]You need to login to perform this action.
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