What about delta Hfsup>o?
For the reaction Cl + e ==> Cl^-, you can look up delta Hof of Cl and for Cl^- and delta H rxn = delta H products - delta H reactants, then convert from mol to 2.65 mg. My text gives +121.7 kJ/mol for deltHof Cl and -226 for Cl^-. Of course that's just delta H; but you could do the same for delta Go.
How much energy is evolved when 2.65 mg of Cl(g) atoms adds electrons to give Cl1-(g) ions.
No clue how to go about solving this problem as it doesn't really relate to any of my examples in my text book or notes. All I can find that related to this problem is electron affinity and E=(K*Q1*Q2)/r where K is a constant. The EA for Cl is -349 kJ/mol. But I just cant find anything else out about this problem. Any help is great as I'm out of ideas.
3 answers
@DrBob222 - I understand up to the part about the conversion of mol to 2.65 mg. I understand that the enthalpy change is 121.3 kJ/mol for Cl(g) and that would be the reactant part of the equation. But whats the deal with working with 2.65mg and the delta Go?
The ionization (is always concerned with the formation of positive ions) of Chlorine is given by the following equation;
Cl --> Cl+ + e- IE = 1251.19 KJ/mol
While the electron affinity (is always concerned with the formation of negative ions) of Chlorine is given by the following equation;
Cl + e- --> Cl- EA = -349 kJ/mol
# moles of Cl = 0.00265 g/35.5 g/mol
= 0.0000746479 mol
Therefore, the energy involved in the reaction is E = EA x nCl
= -0.026 kJ
Exothermic reaction because the energy is released when the atom gains electrons
Cl --> Cl+ + e- IE = 1251.19 KJ/mol
While the electron affinity (is always concerned with the formation of negative ions) of Chlorine is given by the following equation;
Cl + e- --> Cl- EA = -349 kJ/mol
# moles of Cl = 0.00265 g/35.5 g/mol
= 0.0000746479 mol
Therefore, the energy involved in the reaction is E = EA x nCl
= -0.026 kJ
Exothermic reaction because the energy is released when the atom gains electrons
1 answer