# 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.... ?

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

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

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The first thing to do here is to calculate the number of *moles* of hydrogen gas, **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 = (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 *homolysis*, i.e. when each of the two hydrogen atoms gets to **keep** the electron that they contributed to the

Bond-dissociation energies are usually given in *kilojoules per mole*. Let's say that

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 **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 **times smaller** than the bond-dissociation energy of the

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 twosig figs