A) \[-17.297\text{ }kcal\]
B) +17.297 kcal
C) \[-172.97\text{ }kcal\]
D) +172.97 kcal
Correct Answer: A
Solution :
Key Idea: Write the equation for heat of formation of\[C{{H}_{4}}\]first to calculate\[{{n}_{g}}\]then calculate\[\Delta E\](heat of formation of\[C{{H}_{4}}\]at constant volume) by using formula \[\Delta E=\Delta H+\Delta {{n}_{g}}RT\] \[\Delta H=-17.890\text{ }kcal\] Given, \[R=1.987\text{ }kcal\text{ }\deg re{{e}^{-1}}mo{{l}^{-1}}\] \[T=25{}^\circ C=25+273=298K\] \[C(s)+2{{H}_{2}}(g)\xrightarrow{{}}C{{H}_{4}}(g)\] \[\Delta {{n}_{g}}=\]number of moles of gaseous products - number of moles of gaseous reactants \[=1-2=-1\] \[\Delta E=\Delta H-nRT\] \[=-\text{ }17.890+(-1\times 1.987\times {{10}^{3}}\times 298)\] \[=-17.297\text{ }kcal\]
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