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

Question

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 (A)  +250 kJ  (B)  -250 kJ  (C)  +125 kJ  (D)  -125 kJ

Tardigrade Admin · Accepted Answer

CH 2= CH 2+ H 2 longrightarrow CH 3- CH 3 Δ H=(B E) text reactants -(B E) text products =4(B E)C-H+(B E)C=C+(B E)H-H -[6(B E)C-H+(B E)C-C] =-125 kJ

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 $435 k J m o l^{- 1}$ , the value of enthalpy change for the reaction $H_{2} C = C H_{2} (g) + H_{2} (g) ⟶ H_{3} C - C H_{3} (g)$ at $298 K$ will be

$+ 250 k J$

$- 250 k J$

$+ 125 k J$

$- 125 k J$

$C H_{2} = C H_{2} + H_{2} ⟶ C H_{3} - C H_{3}$
$Δ H = (BE)_{reactants} - (BE)_{products}$
$= 4 (BE)_{C - H} + (BE)_{C = C} + (BE)_{H - H}$
$- [6 (BE)_{C - H} + (BE)_{C - C}]$
$= - 125 k J$

Q. If at 298K the bond energies of C−H,C−C,C=C and H−H bonds are respectively 414,347,615 and 435kJmol−1, the value of enthalpy change for the reaction H2​C=CH2​(g)+H2​(g)⟶H3​C−CH3​(g) at 298K will be

+250kJ

−250kJ

+125kJ

−125kJ