## 2 HBr (aq) + 1 Na2CO3 (s) → 2 NaBr (aq) + H2O (ℓ) + 1 CO2 (g)

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## Reaction Type:

Double Displacement/Decomposition of a Carbonate

## 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.

 HBr            Mass: g or Solution Volume: mL of Concentration: mol/L Na2CO3         Mass: g NaBr           Mass: g H2O            Mass: g CO2            Mass: g or Gas Volume: L Heat Released: kJ

## Enthalpy of Reaction

[2ΔHf(NaBr (aq)) + 1ΔHf(H2O (ℓ)) + 1ΔHf(CO2 (g))] - [2ΔHf(HBr (aq)) + 1ΔHf(Na2CO3 (s))]
[2(-361.65) + 1(-285.83) + 1(-393.51)] - [2(-121.55) + 1(-1130.94)] = -28.5999999999999 kJ
-28.60 kJ     (exothermic)

## Entropy Change

[2ΔSf(NaBr (aq)) + 1ΔSf(H2O (ℓ)) + 1ΔSf(CO2 (g))] - [2ΔSf(HBr (aq)) + 1ΔSf(Na2CO3 (s))]
[2(141.42) + 1(69.91) + 1(213.68)] - [2(82.42) + 1(135.98)] = 265.61 J/K
265.61 J/K     (increase in entropy)

## Free Energy of Reaction (at 298.15 K)

From ΔGf° values:
[2ΔGf(NaBr (aq)) + 1ΔGf(H2O (ℓ)) + 1ΔGf(CO2 (g))] - [2ΔGf(HBr (aq)) + 1ΔGf(Na2CO3 (s))]
[2(-365.87) + 1(-237.18) + 1(-394.38)] - [2(-103.97) + 1(-1047.67)] = -107.69 kJ
-107.69 kJ     (spontaneous)

From ΔG = ΔH - TΔS:
-107.79 kJ     (spontaneous)

## Equilibrium Constant, K (at 298.15 K)

7.3709172778e+018
This process is favorable at 25°C.