How to find enthalpy change of reaction from enthalpy change of solution?
The enthalpy change of solution of Na2CO3 is -35 KJ/mol
The enthalpy change os solution of Na2CO3.10H2O is -60 KJ/mol
Use the above data to find the enthalpy change of reaction for the following reaction:
Na2CO3 (s) + 10H2O (l) ---> Na2CO3.10H2O
I have absolutely no idea what to do. Help would be greatly appreciated.
The enthalpy change os solution of Na2CO3.10H2O is -60 KJ/mol
Use the above data to find the enthalpy change of reaction for the following reaction:
Na2CO3 (s) + 10H2O (l) ---> Na2CO3.10H2O
I have absolutely no idea what to do. Help would be greatly appreciated.
Original post by ihateocr12345
The enthalpy change of solution of Na2CO3 is -35 KJ/mol
The enthalpy change os solution of Na2CO3.10H2O is -60 KJ/mol
Use the above data to find the enthalpy change of reaction for the following reaction:
Na2CO3 (s) + 10H2O (l) ---> Na2CO3.10H2O
I have absolutely no idea what to do. Help would be greatly appreciated.
The enthalpy change os solution of Na2CO3.10H2O is -60 KJ/mol
Use the above data to find the enthalpy change of reaction for the following reaction:
Na2CO3 (s) + 10H2O (l) ---> Na2CO3.10H2O
I have absolutely no idea what to do. Help would be greatly appreciated.
Both dissolution processes produce the same outcome (a solution of sodium carbonate)
Construct a Hess cycle with your unknown equation and the two dissolution equations.
Original post by charco
Both dissolution processes produce the same outcome (a solution of sodium carbonate)
Construct a Hess cycle with your unknown equation and the two dissolution equations.
Construct a Hess cycle with your unknown equation and the two dissolution equations.
I'm really confused as I've only ever done this with enthalpy of formation and combustion. Could you help with this please? thanks
Original post by ihateocr12345
I'm really confused as I've only ever done this with enthalpy of formation and combustion. Could you help with this please? thanks
Check out this interactive on Hess law
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