Question

‎50.mL of 0.20M lead (II) nitrate solution, Pb(NO3)2(aq), at 19.6C was added to 30mL of a solution containing excess potassium iodide, KI(aq) also at 19.6C.
The solutions reacted to form a yellow lead (II) iodide precipitate, PbI2(s), and the temperature of the reaction mixture increased to 22.2C.

Calculate the enthalpy change per mole of lead (II) nitrate for the reaction.

Answer 1

How do we know if the reaction is endothermic or exothermic?

Answer 2

Second question first, since it's easier: The temperature of the reaction increased, so energy must have been released. That tells us whether it is endothermic or exothermic.

First question: The total amount of energy released will be equal to:\[Q = m*C{_P}* \Delta T\] where m is the total mass of water, C is the specific heat of water (4.18J/gC) and DT is the temperature change. The energy released PER MOLE of Pb(NO3)2 will be equal to the total energy change divided by the number of moles present in the lead (II) nitrate solution you started with:\[Q/mol = \frac{Q}{0.050L * (\frac{0.20mol}{1L})}\]

Answer 3

Thank you :)

Answer 4

Final answer I got is -85kJ.

Answer 5

-87*