100.00 mL of 0.100 M Pb(NO3)2 and 100.00 mL of 0.200 M HCl are placed in a calorimeter.

Question

Pb^{2+ (aq) + 2 Cl- (aq) ---> PbCl}_2(s).

Both solutions were intially at 27.5^{0 C.} After the reaction, the final temperature was 24.8⁰ C. Calculate the heat of this reaction in KJ/mol of PbCl₂ formed. Assume that the combined solution has a mass of 200.0 g and a specific heat capacity of 4.18 J/g.C_

Answer 1

The formula for finding heat transferred is q = mC∆T

q = heat = ?

m = mass = 200 g

C = specific heat = 4.18 J/gC

∆T = change in temperature = 2.7 degrees

q = (200 g)(4.18 J/g/degree)(2.7 degrees) = 2257 J but notice that the temperature decreased from 27.7 to 24.8 degrees. This indicates that the reaction is ENDOTHERMIC and so q will be positive, i.e. +2257 J

Next, we find the moles of Pb(NO₃)₂ and moles of HCl that are present:

mmoles Pb(NO₃)₂ = 100.00 ml x 0.100 mmol/ml = 10.00 mmoles Pb(NO₃)₂ = 0.01 moles

mmoles HCl = 100.00 ml x 0.200 mmol/ml = 20 mmoles HCl = 0.02 moles

Neither reactant is in limiting supply since it takes 2 moles HCl for each 1 mole of Pb(NO₃)₂

From the balanced equation, 0.01 moles of PbCl₂ will be formed

∆H/mole PbCl₂ = 2257 J/0.01 mol PbCl₂ = 225,700 J/mol = +225.70 kJ/mol