Question

$C_2{H}_4\left(g\right)+H_2\left(g\right)\to C_2{H}_6\left(g\right)$

Using the bond enthalpy data given below, calculate the enthalpy change for the reaction.

Bond	$C-C$	$C=C$	$C-H$	$H-H$
Bond Enthalpy	336.81 kJ / mol	606.68 kJ / mol	410.87 kJ / mol	431.79 kJ / mol

• A. $-$120.08 kJ / mol
• B. $+$120.08 kJ / mol
• C. $-$240.16 kJ / mol
• D. $+$240.16 kJ / mol

Answer 1

The correct option is B $+$120.08 kJ / mol

$H\left(C_2{H}_4\right)=H(C=C)+4\times H(C-H)$

$=606.68+4\times 410.87=2250.16$kJ/mol

Similarly,

$H\left(H_2\right)=431.79$ kJ/mol

and

$H\left(C_2{H}_6\right)=H(C-C)+6\times H(C-H)=336.81+6\times 410.87=2802.03$ kJ/mol

$\therefore$ $H_{\left(reaction\right)}=H\left(C_2{H}_6\right)-\left[H\right(H_2)+H(C_2{H}_4\left)\right]$

$H_{\left(reaction\right)}=2802.03-(431.79+2250.16)=120.08$ kJ/mol.

Hence, the correct option is $B$