7. The ionisation potential of hydrogen is 13.60 eV. Calculate the energy in kJ required to produce 0.1 mole of H ions (given, 1 eV= 96.49 kJ mol-?).
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Answers
Answer:
The ionisation potential may be represented as
The ionisation potential may be represented asH(g)+13.60eV→H⊕(g)+e−
The ionisation potential may be represented asH(g)+13.60eV→H⊕(g)+e−We know 1eV=96.3kJmol−1
The ionisation potential may be represented asH(g)+13.60eV→H⊕(g)+e−We know 1eV=96.3kJmol−113.60eV=96.3×13.60=1309.68kJ
The ionisation potential may be represented asH(g)+13.60eV→H⊕(g)+e−We know 1eV=96.3kJmol−113.60eV=96.3×13.60=1309.68kJThus, energy per mol =1309.68kJ
Answer:
Answer:
The ionisation potential may be represented as
The ionisation potential may be represented asH(g)+13.60eV→H⊕(g)+e−
The ionisation potential may be represented asH(g)+13.60eV→H⊕(g)+e−We know 1eV=96.3kJmol−1
The ionisation potential may be represented asH(g)+13.60eV→H⊕(g)+e−We know 1eV=96.3kJmol−113.60eV=96.3×13.60=1309.68kJ
The ionisation potential may be represented asH(g)+13.60eV→H⊕(g)+e−We know 1eV=96.3kJmol−113.60eV=96.3×13.60=1309.68kJThus, energy per mol =1309.68kJ
Explanation: