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akhil mahajan Grade: 12
        

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..

6 years ago

Answers : (1)

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

6 years ago
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