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The Ionisation potential(IP) values of Al->Al+ + e- is 577.5 KJ/mol and change in enthalpy for Al- > Al3+ +3 e- is 5140 KJ/mol . If 2nd and 3rd IP values are in ratio 2:3 , calculate IP2 and IP3...... plzz give the full solution..
Al-> Al+ + e -- 1 IP 1 Enthalpy change =+577.5 kJ/mol
Al+ -> Al2+ + e --2 IP 2 Let enthalpy change = + 2x kJ/mol
Al2+ -> Al3+ +e --3 IP 3 let enthalpy change = +3x kJ/mol
we have let IP 2 =2x & IP 3 = 3x since their ratio is 2:3 and 2x/3x = 2/3
Adding equation 1,2&3 we get Al->Al3+ + e
enthalpy change = IP 1+IP 2 +IP 3 = 5140kJ/mol
=> 577.5 +2x+3x = 5140
=> 577.5 +5x = 5140
=> 5x= 5140 - 577.5
=> 5x= 4562.5
=> x = 4562.5/5 = 912.5 kJ/mol
Hence IP 2= 2* 912.5= 1825.0 kJ/mol
IP 3= 3*912.5= 2737.5 kJ/mol
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