Question

An insulated bottle contains 1 mole of hydrogen gas at P=1 atm and T=300K. Using a magic wand, you order all covalent bonds in the H2 molecules to break instantly. Assume that the magic wand supplies precisely the amount of energy necessary to br eak the bond in every molecule and makes them chemically inert (so they cant recombine) but does not affect the hydrogen otherwise. When the new equilibrium is established,

a) What is the new temperature of the (now) monoatomic hydrogen?

b) What is the new pressure of the monoatomic hydrogen?

c) Look up the hyd rogen bond dissociation energy to determine how much energy it actually took to perform the magic step. Imagine that the “magic” force keeping the hydrogen atoms inert suddenly fails. If ALL of the molecules were to form again, what would be the new temperature of the (diatomic) gas? (Note: it is interesting to think about what would actually happen once the magic is removed and the atomic hydrogen is allowed to behave according to physicl laws, and what the final state of the system would be. Here, I am not looking for an exact solution (It is pretty clear that It would be hard to convince all hydrogen to convert back to H2, given your result.Do you agree?

Answer 1

We should look at the gas laws to consider the problem when one mole of H2(g) dissociates to form

2 mols of H atoms

H2(g) <-------------> 2H

Accordingly,

At constant T and n ; V is propotional to 1/P

at constant P and n ; V--------------------- T

at constant P & T ; V----------------------n

Thus , V is proportional to nT/ P

or V = R nT/P

Now , a ) T = PV/ Rn

so, initially......n= 1,therefore .......T = PV / R = 300K .......(given )

at equilibria....n = 2 , therefore ...T = 1/2 (PV / R ) = 150 K

Similarly, b) P = RnT / V

so, initially...... P = R(1)T/ V = RT/ V = 1 atm ......(given )

& at equilibria P = R (2) T / V = 2RT / V = 2 atm .

c) Bond dissociation enthalpy of H = 435.8 kJ mol^-1

The new temperature would be 300K , since bon formation is an endothermic process.