When a 6.07-g sample of solid lead(II) nitrate dissolves in 31.9 g of water in a coffee-cup calorimeter (see above figure) the temperature falls from 22.00 oC to 18.47 oC. Calculate H in kJ/mol Pb(NO3)2 for the solution process.
Pb(NO3)2(s) Pb2+(aq) + 2 NO3-(aq)
The specific heat of water is 4.18 J/g-K.
Hsolution = ?kJ/mol
Pb(NO3)2.
Pb(NO3)2(s) ---> Pb2+(aq) + 2 NO3-(aq)
heat absorbed(q) = m*s*DT
m = mass of solution = 31.9+6.07 = 37.97 g
s = specific heat of solution = water = 4.18 j/g.k
DT = 22.00 - 18.47 = 3.53 k
q = 37.97*4.18*3.53
= 560.3 joule
= 0.56 kj
No of mole of Pb(NO3)2 taken = w/m.wt = 6.07/331.2 = 0.0183 mole
DHsol = +q/n
= 0.56/0.0183
= +30.6 kj/mole
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