For Br2, ΔHvap = 31 kJ/mol. If S values for Br2(g) and Br2(l) are 245 J/mol K and 153 J/mol K respectively, what is the normal boiling point for Br2 (l)? I have no idea what to do.
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Hi Salinger!
This question requires you to understand the gibbs free energy equation: ΔG = ΔH - TΔS. Whenever, ΔG is less than zero, the reaction is spontaneous. Whenever ΔG, is greater than zero, the reaction is then not spontaneous or disfavored. However, if ΔG is equal to zero, then neither the forward or reverse reaction is favored and so the reaction is in equilibrium.
Now the key to this question is understanding that during the boiling point of a liquid, it is in equilibrium with the gas it is forming. Thus for this case, ΔG is equal to zero and we have here 0 = ΔH - TΔS where ΔH represents the enthalpy of vaporization, ΔS represents the entropy of vaporization and T represents the normal boiling point for bromine in kelvins. Now we want to solve for the boiling point and so let's rearrange this equation so that we can solve for T:
ΔH = TΔS and so T = ΔH/ΔS
We know our ΔH because it is given (31 kJ/mol) but we don't know ΔS. ΔS is basically the difference in entropy between the final and initial states. Since we know the entropy values of the gas and liquid form (before and after the vaporization), we can simply subtract to find the value of ΔS. 245 - 153 = 92 J/mol K and that is the value of ΔS.
Now plug everything back into the equation we rearranged and solve for the temperature: T = (31 kJ/mol)/(92 J/mol K). NOTE: the energy units are NOT the same. The units for enthalpy was in kilojoules and so we must convert it to joules in order to properly divide the equation. T = (31,000 J/mol)/(92 J/mol K) and so T = 336.96 K but just 340K for significant figures.
Thus, the normal boiling point for Br2(l) is 340 K.
I hope this helped and feel free to ask more questions! :)
This question requires you to understand the gibbs free energy equation: ΔG = ΔH - TΔS. Whenever, ΔG is less than zero, the reaction is spontaneous. Whenever ΔG, is greater than zero, the reaction is then not spontaneous or disfavored. However, if ΔG is equal to zero, then neither the forward or reverse reaction is favored and so the reaction is in equilibrium.
Now the key to this question is understanding that during the boiling point of a liquid, it is in equilibrium with the gas it is forming. Thus for this case, ΔG is equal to zero and we have here 0 = ΔH - TΔS where ΔH represents the enthalpy of vaporization, ΔS represents the entropy of vaporization and T represents the normal boiling point for bromine in kelvins. Now we want to solve for the boiling point and so let's rearrange this equation so that we can solve for T:
ΔH = TΔS and so T = ΔH/ΔS
We know our ΔH because it is given (31 kJ/mol) but we don't know ΔS. ΔS is basically the difference in entropy between the final and initial states. Since we know the entropy values of the gas and liquid form (before and after the vaporization), we can simply subtract to find the value of ΔS. 245 - 153 = 92 J/mol K and that is the value of ΔS.
Now plug everything back into the equation we rearranged and solve for the temperature: T = (31 kJ/mol)/(92 J/mol K). NOTE: the energy units are NOT the same. The units for enthalpy was in kilojoules and so we must convert it to joules in order to properly divide the equation. T = (31,000 J/mol)/(92 J/mol K) and so T = 336.96 K but just 340K for significant figures.
Thus, the normal boiling point for Br2(l) is 340 K.
I hope this helped and feel free to ask more questions! :)
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