1. Tardigrade
  2. Question
  3. Chemistry
  4. If at 298 K the bond energies of C -H, C-C, C=C and H-H bonds are respectively 414,347,615 and 435 kJ mol -1, the value of enthalpy change for the reaction H 2 C = CH 2( g )+ H 2( g ) longrightarrow H 3 C - CH 3( g ) at 298 K will be

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Q. If at $298 K$ the bond energies of $C -H, C-C, C=C$ and $H-H$ bonds are respectively $414,347,615$ and $435 kJ mol ^{-1}$, the value of enthalpy change for the reaction $H _{2} C = CH _{2}( g )+ H _{2}( g ) \longrightarrow H _{3} C - CH _{3}( g )$ at $298 K$ will be

UPSEEUPSEE 2007

Solution:

$CH _{2}= CH _{2}+ H _{2} \longrightarrow CH _{3}- CH _{3}$
$\Delta H=(B E)_{\text {reactants }}-(B E)_{\text {products }}$
$=4(B E)_{C-H}+(B E)_{C=C}+(B E)_{H-H}$
$-\left[6(B E)_{C-H}+(B E)_{C-C}\right]$
$=-125 kJ$