The enthalpy of vaporization of chloroform (CHCl3) is 29.2 kJ mol-1 at its normal boiling point 61.2 Celsius. what is the standard change in entropy for vaporization of chloroform at its normal boiling point?
Answer I got was 29.2/334.2 = 0.0873. (CORRECT?)
You forgot to convert it. The correct answer is 87.3 J/(mol×k).
To calculate the standard change in entropy (ΔS°) for the vaporization of chloroform (CHCl3) at its normal boiling point, we can use the equation:
ΔS° = ΔH°/T
Where ΔH° is the standard enthalpy change for vaporization, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin. We do that by adding 273.15 to the Celsius temperature:
61.2 + 273.15 = 334.35 K
Now, we can substitute the values into the equation:
ΔS° = 29.2 kJ/mol / 334.35 K = 0.0873 kJ/(mol·K)
So, the standard change in entropy for the vaporization of chloroform at its normal boiling point is indeed 0.0873 kJ/(mol·K). Your answer is correct.
To determine the standard change in entropy for the vaporization of chloroform at its normal boiling point, we need to use the equation:
ΔS = ΔH / T
Where:
ΔS is the standard change in entropy
ΔH is the enthalpy of vaporization
T is the temperature in Kelvin
First, let's convert the boiling point from Celsius to Kelvin:
T (Kelvin) = T (Celsius) + 273.15
T (Kelvin) = 61.2 + 273.15
T (Kelvin) = 334.35 K
Now we can substitute the values into the equation:
ΔS = 29.2 kJ/mol / 334.35 K
Calculating this, we get:
ΔS = 0.0873 kJ/(mol·K)
So, the standard change in entropy for the vaporization of chloroform at its normal boiling point is 0.0873 kJ/(mol·K). Therefore, the answer you got, 0.0873, is correct.