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When a 6.07-g sample of solid lead(II) nitrate dissolves in 31.9 g of water in a...

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 deltacap.gifH in kJ/mol Pb(NO3)2 for the solution process.

Pb(NO3)2(s) rtarrow.gif Pb2+(aq) + 2 NO3-(aq)


The specific heat of water is 4.18 J/g-K.



deltacap.gifHsolution = ?kJ/mol Pb(NO3)2.

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Answer #1


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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