repugno
Feb11-04, 05:32 AM
Hello all,
I don't quite understand how to calculate the enthalpy of neutralisation. Example, if I add 50cm^3 of a 1.0 mol dm^-3 solution of HCL to a polystyrene cup. And then add 50cm^3 of a 1.1 mol dm^-3 of NaOH to the same cup, we'll get a reaction with a temperature rise. The temperature rise can be noted and a calculation can be made to find the enthalpy of neutralisation.
So, assuming no heat is lost to the suroundings and the specific heat capacity of the solution is the same as that of water, then
Heat absorbed by solution = m * c * t
= 100 * 4.2 * 6.5
= 2730J
This is the part i don't understand, do I devide this figure with the moles of HCL in the solution or the moles of NaOH?
Any help would be much appreciated. Thank you
I don't quite understand how to calculate the enthalpy of neutralisation. Example, if I add 50cm^3 of a 1.0 mol dm^-3 solution of HCL to a polystyrene cup. And then add 50cm^3 of a 1.1 mol dm^-3 of NaOH to the same cup, we'll get a reaction with a temperature rise. The temperature rise can be noted and a calculation can be made to find the enthalpy of neutralisation.
So, assuming no heat is lost to the suroundings and the specific heat capacity of the solution is the same as that of water, then
Heat absorbed by solution = m * c * t
= 100 * 4.2 * 6.5
= 2730J
This is the part i don't understand, do I devide this figure with the moles of HCL in the solution or the moles of NaOH?
Any help would be much appreciated. Thank you