Question

Heat of atomisation of CH$_4$(g) and C$_2$H$_6$(g) are $x$ kJ/mol and $y$ kJ/mol respectively. Find the maximum wavelength of photon required to dissociate C–C bond in C$_2$H$_6$.

• A. $\dfrac{hcN_A}{y-\frac{3x}{2}}$
• B. $\dfrac{hcN_A}{\frac{4x-6y}{4}}$
• C. $\dfrac{hcN_A}{250\left(\frac{3x}{2}-y\right)}$
• D. $\dfrac{hcN_A}{500(2y-3x)}$

Hint:

Higher bond energy corresponds to shorter wavelength of photon required for bond dissociation.

Answer 1

The Correct Option is D

Step 1: Write atomisation reactions.
\[ CH_4(g) \rightarrow C(g) + 4H(g);\quad \Delta H = x \] \[ C_2H_6(g) \rightarrow 2C(g) + 6H(g);\quad \Delta H = y \]
Step 2: Express bond energies.
\[ x = 4\varepsilon_{C-H} \] \[ y = \varepsilon_{C-C} + 6\varepsilon_{C-H} \]
Step 3: Eliminate $\varepsilon_{C-H$.}
\[ \varepsilon_{C-C} = y - \frac{3x}{2} \]
Step 4: Convert bond energy to photon wavelength.
Energy per mole = $1000\left(y - \frac{3x}{2}\right)$
\[ \lambda_{max} = \frac{hcN_A}{1000\left(y - \frac{3x}{2}\right)} \]
Rewriting:
\[ \lambda_{max} = \frac{hcN_A}{500(2y - 3x)} \]