The heat of atomization of methane and ethane are 360kJ/mol and 620kJ/mol, respectively the longest wavelength light capable of breaking the C−C bond is (Avodrago constant =6.02×1023mol−1).

A
$1.49 \times 10^{3} nm$
B
$2.48 \times 10^{3} nm$
C
$2.48 \times 10^{4} nm$
D
$1.49 \times 10^{4} nm$

Solution:

Given data as

$\begin{array}{l} {CH}_{4} (g) \to C (g) + 4 H (g) Δ H_{1} = 4 ε_{C - H} \\ C_{2} H_{6} (g) \to 2 C (g) + 6 H (g) Δ H_{2} = ε_{C - C} + 6 ε_{C - H} \end{array}$

From $Δ H_{1}$ , we get $ε_{C - H} = (360 / 4) kJ {mol}^{- 1} = 90 kJ {mol}^{- 1}$

From $Δ H_{2}$ , we get $ε_{C - H} = Δ H_{2} - 6 ε_{C - H} = (620 - 6 \times 90) {kJ}^{- 1} = 80 kJ {mol}^{- 1}$

For a single $C - H$ bond, we have

$E = \frac{ε_{C} - H}{N_{A}} = \frac{(80 kJ {mol}^{- 1})}{(6.02 \times 10^{23} {mol}^{- 1})} = 1.33 \times 10^{- 22} kJ = 1.33 \times 10^{- 14} J$

The longest wavelength required to break this bond is

$\begin{array}{l} λ = \frac{h c}{E} = \frac{(6.626 \times 10^{- 34} Js) (3 \times 10^{8} {ms}^{- 1})}{(1.33 \times 10^{- 19} J)} = 1.495 \times 10^{- 6} m \\ = 1.495 \times 10^{3} nm \end{array}$

The heat of atomization of methane and ethane are $360 kJ / mol$ and $620 kJ / mol$ , respectively the longest wavelength light capable of breaking the $C - C$ bond is (Avodrago constant $= 6.02 \times 10^{23} {mol}^{- 1}$ ).

Solution:

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