Answer :
Final answer:
The entropy change, ΔS, for vaporization of 11.0 g ethanol at 79.0°C is calculated using the formula ΔS = q_reversible/T. After converting all quantities to the correct units, we get ΔS = 26.17 J K⁻¹.
Explanation:
In order to calculate the entropy change ΔS, we'll use the formula ΔS = q_reversible/T. The energy (q_reversible) is the heat of vaporization, but it must be in Joules, not kJ. So, we first convert 38.6 kJ to J (1 kJ = 1000 J) giving us 38600 J. Next, we will convert the mass of ethanol to moles using its molar mass. We have 11.0 g of ethanol and using its molar mass as 46.07 g/mol, that gives us 0.239 mol. The ΔH_vap is a molar quantity, so we need to multiply the given ΔH_vap (38.6 kJ/mol) by the number of moles we have (0.239 mol), giving us 9214 J. Now, we have everything to use our formula ΔS = q_reversible/T. The temperature (T) must be in Kelvin, so convert 79.0°C to K by adding 273.15, giving us 352.15 K. This gives us ΔS = 9214 J / 352.15 K = 26.17 J K⁻¹.
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