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Consider the reactions: H 2  2H(I) H  H + + e - (II) The ionization energy of H is 13.53 eV and the degeneracies of the electron and H are 2, while that.

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Presentation on theme: "Consider the reactions: H 2  2H(I) H  H + + e - (II) The ionization energy of H is 13.53 eV and the degeneracies of the electron and H are 2, while that."— Presentation transcript:

1 Consider the reactions: H 2  2H(I) H  H + + e - (II) The ionization energy of H is 13.53 eV and the degeneracies of the electron and H are 2, while that of the proton is 1. It is suggested that the dissociation of H 2 is essentially complete before the ionization of atomic H begins. Verify this following these steps: a.At a total P = 1atm, find T at which [H + ]/[H] = 0.01 b.At this temperature T show that [H]>>[H 2 ]

2 and the equilibrium constants for both reactions are where the volume, V, is At equilibrium

3 Combining the last equations and assuming that the initial number of molecules for H 2 is equal to Avogadro’s number we get a system of three equations and three unknown variables

4 From the last equation, it can be seen that, at that temperature, the reactions do not occur simultaneously. Then, to find the temperature at which we only need to consider the second reaction and assume that Solving these equations, we obtain that.

5 part b Now, to check if the assumption of independent reactions is correct, it is required to solve the equilibrium for the first reaction at the found temperature. Assuming that the initial number of molecules for H 2 is N av, we have Solving this equation we obtain that the number of atoms of hydrogen with respect to the initial number of molecules of hydrogen tends to 2, which means and


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