Chem Calculations (1 Viewer)

[Sunyata]

When 25 mL of a solution at 23.2 degrees Celsius containing 5.00 * 10^(-3) mol NaOH was mixed with 50 mL of a solution also at 23.2 degrees Celsius that contained 6.00 * 10^(-3) mol HCl in a light plastic beaker, the final temp. of the mixture was 24.1 degrees Celsius. Calculate the heat released and hence the enthalpy change per mole for the neutralisaiton reaction. Take the density and specific heat capacity of all solutions involved as 1.00 g/mL and 4.2 J/K/g respectively, assume that the container had negligible heat capacity and that heat losses to the surroundings were negligible,

Man, this problem was so long I'm completely confused.

Please help?

 

[DInfinity]

Ok, I probably got this all messed up, but I had a go at it xD


n is the number of moles reacted
C is the specific heat capacity
T is temerature





Reaction is exothermic, as there was an increase in temperature. Heat released from 1/8000 moles reacted was 4.725 x 10^(-4), so I guess you multiply that value by 8000 to get the enthalpy change per mole. But this probably all looks like BS, like it seems to me xD

 

[Sunyata]

Hmm the answers are meant to be 57 kJ/mol and - 57 kJ/mol