Calculate the molar entropy of vaporization for
liquid hydrogen iodide at its boiling point,
�34.55°C.
HI(l)7HI(g) �Hvap
� 19.76 kJ/mo
Divide the molar heat of vaporization (which you know) by the absolute temperature at which the phase transition occurs, 307.71 K. .
To calculate the molar entropy of vaporization (ΔSvap) for liquid hydrogen iodide (HI) at its boiling point, we can utilize the following formula:
ΔSvap = ΔHvap / T
Where:
ΔSvap is the molar entropy of vaporization,
ΔHvap is the molar enthalpy of vaporization, and
T is the temperature in Kelvin.
First, we need to convert the boiling point of hydrogen iodide from Celsius to Kelvin. The conversion can be done as follows:
T (K) = T (°C) + 273.15
T (K) = 34.55°C + 273.15
T (K) = 307.7 K
Next, we can substitute the given values into the equation:
ΔSvap = 19.76 kJ/mol / 307.7 K
Now, let's calculate ΔSvap:
ΔSvap = 0.0642 kJ/(mol·K)
Therefore, the molar entropy of vaporization for liquid hydrogen iodide at its boiling point is approximately 0.0642 kJ/(mol·K).
To calculate the molar entropy of vaporization for liquid hydrogen iodide (HI) at its boiling point, we need to use the equation:
ΔSvap = ΔHvap / T
where:
- ΔSvap is the molar entropy of vaporization
- ΔHvap is the molar enthalpy of vaporization
- T is the temperature in Kelvin (K)
First, we need to convert the boiling point of hydrogen iodide in Celsius to Kelvin:
T = 34.55°C + 273.15 = 307.70 K
Next, we can substitute the values of ΔHvap and T into the equation:
ΔSvap = 19.76 kJ/mol / 307.70 K
Now, let's do the calculation:
ΔSvap = 0.064 kJ/(mol·K)
Therefore, the molar entropy of vaporization for liquid hydrogen iodide at its boiling point is 0.064 kJ/(mol·K).