Question

Using the data provided, calculate the multiple bond energy $\left(kJmo{l}^{-1}\right)$ of a $C\equiv C$ bond in $C_2{H}_2$. That energy is:

(take the bond energy of a $C-H$ bond as 350 $kJmo{l}^{-1}$)$$\begin{array}{c}2C\left(s\right)+H_2\left(g\right)\to C_2{H}_2\left(g\right);\Delta H=225kJmo{1}^{-1}\\ 2C\left(s\right)\to 2C\left(g\right);\Delta H=1410kJmo{1}^{-1}\\ H_2\left(g\right)\to 2H\left(g\right);\Delta H=330kJmo{1}^{-1}\end{array}$$

• A. $1165$
• B. $837$
• C. $865$
• D. $815$

Answer 1

The correct option is D $815$

$$\begin{array}{c}2C\left(s\right)+H_2\left(g\right)\to C_2{H}_2\left(g\right);\Delta H=225kJmo{1}^{-1}\\ 2C\left(s\right)\to 2C\left(g\right);\Delta H=1410kJmo{1}^{-1}\\ H_2\left(g\right)\to 2H\left(g\right);\Delta H=330kJmo{1}^{-1}\end{array}$$

From the first equation,

$225=\left[\Delta H\right(2C\left(s\right)\to 2C\left(g\right)+\Delta H\left(H_2\right(g)\to 2H(g\left)\right)\left]\right[2\times BE(C-H)+1\times \text{BE}(C\equiv C)]$

Substitute values in the above expression.

$225=[1410+330]-[2\times 350+1\times \text{BE}(C\equiv C\left)\right]$

$225=[1410+330]-[700+\text{BE}(C\equiv C\left)\right]$

$225=1740-700-\text{BE}(C\equiv C)$

$\text{BE}(C\equiv C)=815\text{kJ /mol}$

Thus, the multiple bond energy $\left({\text{kJmol}}^{-1}\right)$ of a $C\equiv C$ bond in $C_2{H}_2$ is $815\text{kJ /mol}$.

Hence, the correct option is $\text{D}$