If at 298K the bond energies of C-H,C-C,C=C and H-H bonds are 414,347,615 and 435kJmol- 1 respectively, the value of enthalpy change for the reaction. (CH)2=(CH)2 (g)+H2 (g) arrow (CH)3-(CH)3 (g)298K will be :

Question

If at 298K the bond energies of C-H,C-C,C=C and H-H bonds are 414,347,615 and 435kJmol- 1 respectively, the value of enthalpy change for the reaction. (CH)2=(CH)2 (g)+H2 (g) arrow (CH)3-(CH)3 (g)298K will be : (A) -250kJ (B) +125kJ (C) -125kJ (D) +250kJ

Tardigrade Admin · Accepted Answer

CH2=CH2+H2 arrow CH3-CH3 triangle H=HR-HP triangle H=615+4× 414+435-[347 + 6 × 414] triangle H=-125kJ

Q. If at $298 K$ the bond energies of $C - H, C - C, C = C$ and $H - H$ bonds are $414, 347, 615$ and $435 k J m o l^{- 1}$ respectively, the value of enthalpy change for the reaction. $(C H)_{2} = (C H)_{2 (g)} + H_{2 (g)} \to (C H)_{3} - (C H)_{3 (g)} 298 K$ will be :

$- 250 k J$

$+ 125 k J$

$- 125 k J$

$+ 250 k J$

$C H_{2} = C H_{2} + H_{2} \to C H_{3} - C H_{3}$
$△ H = H_{R} - H_{P}$
$△ H = 615 + 4 \times 414 + 435 - [347 + 6 \times 414]$
$△ H = - 125 k J$

Q. If at 298K the bond energies of C−H,C−C,C=C and H−H bonds are 414,347,615 and 435kJmol−1 respectively, the value of enthalpy change for the reaction. (CH)2​=(CH)2(g)​+H2(g)​→(CH)3​−(CH)3(g)​298K will be :

−250kJ

+125kJ

−125kJ

+250kJ