Question

it takes 945. kj/mol to break a nitrogen - nitrogen triple bond. calculate the maximum wavelength of light for which a nitrogen - nitrogen triple bond could be broken by absorbing a single photon. round your answer to 3 significant digits.

Explanation:

Step1: Convert energy per mole to energy per photon

First, convert the energy required to break one - mole of nitrogen - nitrogen triple bonds from kJ/mol to J/photon. We know that 1 mole contains $N_A = 6.022\times10^{23}$ photons. The energy per mole is $E_{mol}=945\ kJ/mol = 945\times10^{3}\ J/mol$. So the energy per photon $E$ is $E=\frac{E_{mol}}{N_A}=\frac{945\times 10^{3}\ J/mol}{6.022\times 10^{23}\ mol^{-1}}$.
$E=\frac{945\times10^{3}}{6.022\times10^{23}}\ J\approx1.57\times 10^{-18}\ J$

Step2: Use the energy - wavelength relationship

The energy of a photon is given by the formula $E = h
u=\frac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\ J\cdot s$ is Planck's constant, $c = 3.00\times10^{8}\ m/s$ is the speed of light, and $\lambda$ is the wavelength of the photon. We can re - arrange the formula to solve for $\lambda$: $\lambda=\frac{hc}{E}$.
Substitute $h = 6.626\times10^{-34}\ J\cdot s$, $c = 3.00\times10^{8}\ m/s$, and $E = 1.57\times10^{-18}\ J$ into the formula.
$\lambda=\frac{6.626\times 10^{-34}\ J\cdot s\times3.00\times 10^{8}\ m/s}{1.57\times 10^{-18}\ J}$
$\lambda=\frac{19.878\times10^{-26}}{1.57\times10^{-18}}\ m\approx1.27\times10^{-7}\ m$

Step3: Convert the wavelength to nanometers

Since $1\ m = 10^{9}\ nm$, then $\lambda=1.27\times10^{-7}\ m\times10^{9}\ nm/m = 127\ nm$

Answer:

127 nm