A) \[-\text{ }17900\text{ cal}\]
B) \[\text{ }\!\!~\!\!\text{ 17900 cal}\]
C) \[\text{17308 cal}\]
D) \[~-\text{ }17308\text{ cal}\]
Correct Answer: D
Solution :
\[C(graphite)+2{{H}_{2}}(g)\xrightarrow{{}}C{{H}_{4}}(g);\] \[\Delta H=-17900\,cal\] \[\Delta E=?\] \[\Delta H=\Delta E+\Delta {{n}_{g}}RT\] where,\[\Delta {{n}_{g}}=\]number of moles of gaseous products - number of moles of gaseous reactants \[=1-2\] \[=-1\] \[\therefore \] \[\Delta H=\Delta E-1\times R\times T\] \[-17900=\Delta \Epsilon -1\times 298\times 2\] \[\Delta \Epsilon =-17900+596\] \[\therefore \] \[\Delta \Epsilon =-17304\,\text{cal}\]
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