Urgent enthalpy question (1 Viewer)

[clintmyster]

50mL 0.20mol/L lead Nitrate solution is added to excess potassium iodide contained in 30mL of solution. Both solutions are initiallly at 19.6 degrees. After mixing, the temperature rose to 22.2 degrees. Calculate the enthalpy change (per mole of Pb2+) for the reaction?

Answer is : -87kJ/mol

what do you do

 

[minijumbuk]

ΔH= -mCΔT
Since PI is in excess, we do not need to take into account the molarity of PI. But since it's in 30 mL solution, it adds onto the heat sent into the solution. Thus, the mass of the solution is 50mL + 30mL. For calculation purposes, take 1 mL = 1 gram (They will never expect you to deal with density of solution etc)

So, the value of 'm' here is 80 g. C is the specific heat capacity. Since it specifically says "solution", it means that the solution is in water, and thus has a value of 4.18. For ΔT, simply subtract intial temperature from final temperature.

So, ΔH= -80 x 4.18 x (22.2-19.6)
=

Now, the moles of Pb2+ ions in the solution can be calculated by n=CV
Here, n=0.05 x 0.2
= 0.01

So ΔH_{molar[/su] = /0.01
= -87kJ/mol.}