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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..`
7 years ago

Avinash kumar Bharti
32 Points
```										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
```
7 years ago
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