A) \[-372\text{ }kcal\]
B) \[162\text{ }kcal\]
C) \[-240\text{ }kcal\]
D) \[183.5\text{ }kcal\]
Correct Answer: A
Solution :
Combustion equation for ethane is : \[{{C}_{2}}{{H}_{6}}(g)+\frac{7}{2}{{O}_{2}}(g)\to 2C{{O}_{2}}(g)+3{{H}_{2}}O(g)\] \[\Delta {{H}^{o}}=?\] \[\Delta {{H}^{o}}=\{\Sigma \Delta H_{f}^{o}\,products\}-\{\Sigma \Delta H_{f}^{o}\,reac\tan ts\}\] \[=\{2\Delta H_{f}^{o}\,(C{{O}_{2}})-3\Delta {{H}_{f}}({{H}_{2}}O)\}\] \[-\{\Delta H_{f}^{o}\,({{C}_{2}}{{H}_{6}})+7/2\Delta H_{f}^{o}({{O}_{2}})\}\] \[=\{2\times (-94.1)+3\times (-68.3)\}\] \[-\{(-21.1)+7/2(O)\}\] \[=\{-188.2-204.9\}-\{-21.1\}\] \[=-\text{ }393.1+21.1\] \[=-372kcal\]You need to login to perform this action.
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