Questions: What is the wavelength of light that would be required to perform the cis-trans isomerization of one molecule of 2-butene? λ= nm TOOLS × 10^y Which region of the electromagnetic spectrum is this wavelength of light within? ultraviolet visible X-ray infrared

Solution Steps

The cis-trans isomerization of 2-butene involves breaking and reforming a pi bond. The energy required for this process is typically around 250 kJ/mol. This energy corresponds to the energy of the photons needed to induce the isomerization.

The energy of a single photon can be calculated using the formula: \[ E = \frac{E_{\text{molecule}}}{N_A} \] where \( E_{\text{molecule}} = 250 \, \text{kJ/mol} = 250,000 \, \text{J/mol} \) and \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \) is Avogadro's number.

\[ E = \frac{250,000 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} = 4.150 \times 10^{-19} \, \text{J} \]

The energy of a photon is related to its wavelength by the equation: \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \, \text{J s} \) is Planck's constant and \( c = 3.00 \times 10^8 \, \text{m/s} \) is the speed of light.

Rearranging for wavelength, we have: \[ \lambda = \frac{hc}{E} \]

Substituting the values: \[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J s} \times 3.00 \times 10^8 \, \text{m/s}}{4.150 \times 10^{-19} \, \text{J}} \]

\[ \lambda = 4.796 \times 10^{-7} \, \text{m} = 479.6 \, \text{nm} \]

The wavelength of 479.6 nm falls within the visible region of the electromagnetic spectrum, which ranges from approximately 380 nm to 750 nm.

Final Answer