A) \[-17kcal\]
B) \[-111kcal\]
C) \[-170kcal\]
D) \[-85kcal\]
Correct Answer: A
Solution :
For reaction, \[C(s)+2{{H}_{2}}(g)\xrightarrow{{}}C{{H}_{4}}(g),\,\,\Delta H{}^\circ =?\] \[\Delta H{}^\circ =-\][(\[\Delta H{}^\circ \]of combustion of\[C{{H}_{4}}\])\[-\](\[\Delta H{}^\circ \]of combustion of\[C+2\times \Delta H\]of combustion of\[{{H}_{2}}\])] \[C+{{O}_{2}}\to 2{{H}_{2}}O;\,\,\Delta H=-94kcal\] ... (i) \[2{{H}_{2}}+{{O}_{2}}\to 2{{H}_{2}}O;\,\,\Delta H=-68\times 2kcal\] ... (ii) \[C{{H}_{4}}+2{{O}_{2}}\to C{{O}_{2}}+2{{H}_{2}}O;\,\,\Delta H=-213kcal\]... (iii) \[\Delta H{}^\circ =[(-213)-(-94+2\times -68)]kcal/mol\] \[=-[-213+230]=-17kcal/mol\]
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