Question

Calculate the enthalpy of the following reaction. $H_{2}C=C{H}_2\left(g\right)+H_2\left(g\right)\to C{H}_3-C{H}_3\left(g\right)$

The bond energies of C - H, C - C, C = C & H - H are 99, 83, 147 & 104 Kcal respectively.

• A. 30 K cal
• B. -30 K cal
• C. 20 K cal
• D. -20 K cal

Answer 1

The correct option is B

-30 K cal

The reaction given to us is

$\underset{\stackrel{|}{H}}{\stackrel{\underset{|}{H}}{C}}=\underset{\stackrel{|}{H}}{\stackrel{\underset{|}{H}}{C\left(g\right)}}+H-H\left(g\right)⟶H-\underset{\stackrel{|}{H}}{\stackrel{\underset{|}{H}}{C}}-\underset{\stackrel{|}{H}}{\stackrel{\underset{|}{H}}{C}}-H\left(g\right);△H=?$

$△$ H = sum of bond energies of reactants - sum of bond energies of products
=$[△H_{C=C}+4△H_{C-H}+△H_{H-H}]-[△H_{C-C}+6\times △H_{C-H}]$

= (147 + 4 × 99 + 104) - (83 + 6 × 99)

$\therefore △H=-30Kcal$