Question

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?

Answer 1

Here's what I got.

Explanation:

The first thing to do here is to calculate the number of moles of hydrogen gas, `"H"_2`, present in your sample by using the ideal gas law equation.

`color(blue)(ul(color(black)(PV = nRT)))`

Keep in mind that you have

• `R = 0.0821("atm" * "L")/("mol" * "K") ->` the universal gas consatnt
• `P = "100,000"/"101,325"color(white)(.)"atm " ->` since `"1 bar = 100,000 Pa"`
• `T = "300.15 K " ->` since `0^@"C" = "273.15 K"`

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

`n = (PV)/(RT)`

`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 `"H"-"H"` bond by homolysis, i.e. when each of the two hydrogen atoms gets to keep the electron that they contributed to the `"H"-"H"` bond.

Bond-dissociation energies are usually given in kilojoules per mole. Let's say that `BDE` `"kJ mol"^(-1)` represents the bond-dissociation energy of the `"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)` `"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 `"H"-"H"` bond.

You can thus say that

`E_ "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

`BDE color(white)(.)["kJ mol"^(-1)]= 436`

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

This will get you

`E_"total" = 0.07005 * "436 kJ" = "30.54 kJ " ->` rounded to two sig figs