Use four significant figures please.
For the decomposition of barium carbonate, consider the following thermodynamic data: Calculate the temperature in kelvins above which this reaction is spontaneous. Calculate the equilibrium constant for the following reaction at room temperature, 25 degree C: BaCO3(s) rightarrow BaO(s) + CO2(g) Express your answer numerically to three significant figures. Barium oxide, BaO, is manufactured by decomposition of barium carbonate, BaCO3, in a furnace: BaCO3(s) rightarrow BaO(s) + CO2(g) The heat produced by a furnace maintained at a temperature of 600 K can decompose barium carbonate to barium oxide. The spontaneity of a reaction can be determined from the free energy change for the reaction, Delta G degree degree.
The Correct Answer and Explanation is:
To determine when the decomposition of barium carbonate is spontaneous, we use the Gibbs free energy equation: ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ
A reaction becomes spontaneous when: ΔG∘<0\Delta G^\circ < 0
So we solve for the temperature TT at which: ΔG∘=0⇒T=ΔH∘ΔS∘\Delta G^\circ = 0 \Rightarrow T = \frac{\Delta H^\circ}{\Delta S^\circ}
Step 1: Convert units
- ΔH∘=243.5 kJ/mol=243500 J/mol\Delta H^\circ = 243.5 \, \text{kJ/mol} = 243500 \, \text{J/mol}
- ΔS∘=172.0 J/mol\cdotpK\Delta S^\circ = 172.0 \, \text{J/mol·K}
Step 2: Calculate the temperature
T=243500 J/mol172.0 J/mol\cdotpK=1415.7 KT = \frac{243500 \, \text{J/mol}}{172.0 \, \text{J/mol·K}} = 1415.7 \, \text{K}
✅ Answer:
1415 K (to four significant figures: 1415.698 K)
Part 2: Calculate the equilibrium constant KK at 25°C (298 K)
We use: ΔG∘=−RTlnK⇒lnK=−ΔG∘RT\Delta G^\circ = -RT \ln K \Rightarrow \ln K = -\frac{\Delta G^\circ}{RT}
We first compute ΔG∘\Delta G^\circ at 298 K: ΔG∘=ΔH∘−TΔS∘=243500−(298)(172.0)=243500−51256=192244 J/mol\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ = 243500 – (298)(172.0) = 243500 – 51256 = 192244 \, \text{J/mol}
Now, using R=8.314 J/mol\cdotpKR = 8.314 \, \text{J/mol·K}: lnK=−192244(8.314)(298)=−77.46⇒K=e−77.46≈1.00×10−34\ln K = -\frac{192244}{(8.314)(298)} = -77.46 \Rightarrow K = e^{-77.46} \approx 1.00 \times 10^{-34}
✅ Equilibrium Constant at 25°C:
K=1.00×10−34K = 1.00 \times 10^{-34}
Explanation
The spontaneity of a chemical reaction depends on the sign of the Gibbs free energy change, ΔG∘\Delta G^\circ. For the decomposition of barium carbonate (BaCO3(s)→BaO(s)+CO2(g)\text{BaCO}_3(s) \rightarrow \text{BaO}(s) + \text{CO}_2(g)), we are provided with the enthalpy change (ΔH∘=243.5 kJ/mol\Delta H^\circ = 243.5 \, \text{kJ/mol}) and the entropy change (ΔS∘=172.0 J/mol\cdotpK\Delta S^\circ = 172.0 \, \text{J/mol·K}).
Spontaneity is achieved when ΔG∘<0\Delta G^\circ < 0. Setting ΔG∘=0\Delta G^\circ = 0 allows us to calculate the threshold temperature above which the reaction becomes spontaneous. Converting enthalpy to joules and dividing by the entropy gives T=243500172.0=1415.698 KT = \frac{243500}{172.0} = 1415.698 \, \text{K}. Thus, the decomposition is spontaneous at temperatures above approximately 1415.7 K.
To further analyze the reaction’s behavior under standard conditions (room temperature, 25°C or 298 K), we use the Gibbs free energy change to calculate the equilibrium constant KK. A positive ΔG∘\Delta G^\circ at this temperature indicates the reaction is non-spontaneous under standard conditions. The computed equilibrium constant, K=1.00×10−34K = 1.00 \times 10^{-34}, is extremely small, confirming that at room temperature, barium carbonate hardly decomposes.
This helps explain why a furnace (at around 600 K) is used industrially to drive this decomposition. Although 600 K is below the threshold for spontaneity, external energy helps shift the reaction toward products. The calculations illustrate the strong temperature dependence of reaction spontaneity and equilibrium in thermodynamically controlled processes
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