# The energy required for the excitation of H atom its ground state to the 2nd excited state is 2.67 times smaller than the dissociation energy of H_2(g). If H_2 placed in 1.0 L flask at 27^@ C and 1.0 bar is ......continued.... ?

## .... to be excited to their 2nd excited state, What will be the total energy consumption?

Jul 14, 2017

Here's what I got.

#### Explanation:

The first thing to do here is to calculate the number of moles of hydrogen gas, ${\text{H}}_{2}$, present in your sample by using the ideal gas law equation.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Keep in mind that you have

• $R = 0.0821 \left(\text{atm" * "L")/("mol" * "K}\right) \to$ the universal gas consatnt
• $P = \text{100,000"/"101,325"color(white)(.)"atm } \to$ since $\text{1 bar = 100,000 Pa}$
• $T = \text{300.15 K } \to$ since ${0}^{\circ} \text{C" = "273.15 K}$

Rearrange to solve for $n$ and plug in your values to find

$n = \frac{P V}{R T}$

n = ("100,000"/"101,325" color(red)(cancel(color(black)("atm"))) * 1.0 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 300.15color(red)(cancel(color(black)("K")))) = "0.04005 moles H"_2

Now, the bond-dissociation energy for hydrogen gas tells you the change in enthalpy that occurs when you split a $\text{H"-"H}$ bond by homolysis, i.e. when each of the two hydrogen atoms gets to keep the electron that they contributed to the $\text{H"-"H}$ bond. Bond-dissociation energies are usually given in kilojoules per mole. Let's say that $B D E$ ${\text{kJ mol}}^{- 1}$ represents the bond-dissociation energy of the $\text{H"-"H}$ bond.

You can say that you will need

0.04005 color(red)(cancel(color(black)("moles H"_2))) * (BDEcolor(white)(.)"kJ")/(1color(red)(cancel(color(black)("mole H"_2)))) = (0.04005 * BDE) $\text{kJ}$

of energy in order to convert $0.04005$ moles of hydrogen gas to

0.04005 color(red)(cancel(color(black)("moles H"_2))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2)))) = "0.0801 moles H"

At this point, you know that the energy required the electron present in a hydrogen atom to the second excited state is $2.67$ times smaller than the bond-dissociation energy of the $\text{H"-"H}$ bond.

You can thus say that

${E}_{\text{2nd excited state" = (BDEcolor(white)(.)"kJ mol"^(-1))/2.67 = (BDE)/2.67color(white)(.)"kJ mol}}^{- 1}$

This means that you will need

0.0801 color(red)(cancel(color(black)("moles H"))) * (((BDE)/2.67)color(white)(.)"kJ")/(1color(red)(cancel(color(black)("mole H")))) = (0.03 * BDE)color(white)(.)"kJ"

of energy to excited all the hydrogen atoms present in your sample.

The total amount of energy required will thus be

E_"total" = overbrace((0.04005 * BDE)color(white)(.)"kJ")^(color(blue)("needed to split all the H"_2color(white)(.)"molecules")) + overbrace((0.03 * BDE)color(white)(.)"kJ")^(color(blue)("needed to excite all the H atoms"))

E_"total" = color(darkgreen)(ul(color(black)((0.07005 * BDE)color(white)(.)"kJ")))

For a numerical solution, you can take

$B D E \textcolor{w h i t e}{.} \left[{\text{kJ mol}}^{- 1}\right] = 436$

https://en.wikipedia.org/wiki/Bond-dissociation_energy

This will get you

${E}_{\text{total" = 0.07005 * "436 kJ" = "30.54 kJ }} \to$ rounded to two sig figs