Discussion Forum

1)

Using the data provided, calculate the multiple bond energy (kJ mol^-1)of a $C\equiv C$ bond $C_{2}H_{2}$. That energy is (take the bond energy of a C-H bond as 350k Jmol-1)

$2C(s)+H_{2}(g) \rightarrow   C_{2}H_{2}(g);$     $\triangle H=225kJ mol^{-1}$

$2C(s) \rightarrow  2 C (g)$      $\triangle H=1410 kJ mol^{-1}$

$H_{2}(g) \rightarrow 2H(g)$   $\triangle H=330 kJ Mol^{-1}$

Answer:

Option D

Explanation:

For the calaculation of $C\equiv C$  bond energy we must first calculate dissociation energy of $C_{2}H_{2}$ as

 $C_{2}H_{2}$(g)$ \rightarrow$ $2C(g)+2H(g)$.....(i)

 Using the given bond energoes and enthalpies

 $C_{2}H_{2}(g) \rightarrow 2C(g)+2H(g);$

                             $\triangle H=-225kJ$ ........(ii)

$2C (s) \rightarrow 2C(g);$ $ \triangle H=1410kJ$..........(iii)

 $H_{2}(g) \rightarrow 2H(g);$ $\triangle H=330kJ$......(iv)

 adding equation (ii),(iii) and (iv) gives equation(i)

 $\Rightarrow$      $C_{2}H_{2}(g) \rightarrow 2C(g)+2H(g)$;

                                       $\triangle H=1515 kJ$

 $\Rightarrow$     1515kJ=$2 \times$ (C-H)BE+($C\equiv C$)BE

 =$2 \times 350$+($C\equiv C$)BE

$\Rightarrow$     $(C\equiv C)$BE=1515-700=815 kJ/mol