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