## Pb(NO3)2 (aq) + 2 NaBr (aq) → PbBr2 (s) + 2 NaNO3 (aq)

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## Reaction Type:

Double Displacement

## Stoichiometry

Enter a mass or volume in one of the boxes below. Upon hitting submit, the stoichiometric equivalents will be calculated for the remaining reactants and products. All gases are assumed to be at STP.

 Pb(NO3)2       Mass: g or Solution Volume: mL of Concentration: mol/L NaBr           Mass: g or Solution Volume: mL of Concentration: mol/L PbBr2          Mass: g NaNO3          Mass: g Heat Released: kJ

## Enthalpy of Reaction

[1ΔHf(PbBr2 (s)) + 2ΔHf(NaNO3 (aq))] - [1ΔHf(Pb(NO3)2 (aq)) + 2ΔHf(NaBr (aq))]
[1(-277.4) + 2(-447.46)] - [1(-416.42) + 2(-361.65)] = -32.6000000000001 kJ
-32.60 kJ     (exothermic)

## Entropy Change

[1ΔSf(PbBr2 (s)) + 2ΔSf(NaNO3 (aq))] - [1ΔSf(Pb(NO3)2 (aq)) + 2ΔSf(NaBr (aq))]
[1(161.13) + 2(205.44)] - [1(303.38) + 2(141.42)] = -14.21 J/K
-14.21 J/K     (decrease in entropy)

## Free Energy of Reaction (at 298.15 K)

From ΔGf° values:
[1ΔGf(PbBr2 (s)) + 2ΔGf(NaNO3 (aq))] - [1ΔGf(Pb(NO3)2 (aq)) + 2ΔGf(NaBr (aq))]
[1(-260.75) + 2(-373.24)] - [1(-247.08) + 2(-365.87)] = -28.41 kJ
-28.41 kJ     (spontaneous)

From ΔG = ΔH - TΔS:
-28.36 kJ     (spontaneous)

## Equilibrium Constant, K (at 298.15 K)

94949.70289
This process is favorable at 25°C.

## Reference(s):

Ebbing, Darrell D. General Chemistry 3rd ed.; Houghton Mifflin Company: Boston, MA, 1990; p 98.