A quantity of 200 mL of 0.862M HCLis mixed with 200mL of 0.431M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCL and Ba(OH)2 solutions is the same at 20.48C. For the process
H+(aq) + OH- ---------> H2O(l)
the heat of neutralization is -56.2kJ/mol. What is the final temperature of the mixed solution?
How to calcula the enthalpy change in different M?? 作者: 一人一個 時間: 2009-2-19 13:38
Steps to solve the question:
(1) find the number of moles of HCl and Ba(OH)2 in the mixture
(you should find that these numbers are the same)
(2) find the number of moles of H+ and OH- ions (by mole ratio)
(3) given the enthalpy change for neutralization, multiply this value with the number of moles of H2O to be formed to find the total amount of energy released
(4) since you know the specific heat capacity of water, you can work out the temperature change by the equation Q = mcT (the trick is: what is the volume of the mixture? 200ml or 400ml?)
Don't get confused by the various molarities of HCl and Ba(OH)2 solutions. Always convert concentrations to numbers of moles before doing any calculations. Mole is the most basic and most convenient units in this type of calculations. 作者: 歡樂屋仔 時間: 2009-2-19 18:47
