Calculate the molar entropy of vaporization for liquid hydrogen iodide at its boiling point -34.55 degrees celcius. HI(l) <=> HI(g)

deltaHvap = 19.76 kJ/mol

delta Hvap = Tvap x delta Svap

Substitute and solve for delta Svap. Don't forget Tvap must be in kelvin. Note the correct spelling of celsius.

didn't help me much :? anyone else?

Why did the thermometer go to therapy?

Because it had a boiling point identity crisis!

But let's get back to your question. To calculate the molar entropy of vaporization (ΔSvap) for liquid hydrogen iodide (HI) at its boiling point, we need to use the equation:

ΔSvap = ΔHvap / T

Where ΔHvap is the heat of vaporization and T is the temperature in Kelvin.

Since the boiling point of HI is -34.55 degrees Celsius, we need to convert it to Kelvin:

T = -34.55 + 273.15 = 238.6 K

Now we can plug in the values:

ΔHvap = 19.76 kJ/mol = 19760 J/mol

ΔSvap = 19760 J/mol / 238.6 K

Calculating this gives us the molar entropy of vaporization for liquid HI at its boiling point.

To calculate the molar entropy of vaporization (ΔSvap) for liquid hydrogen iodide (HI), we need to use the following equation:

ΔSvap = ΔHvap / T

Where:
ΔHvap is the molar enthalpy of vaporization (given as 19.76 kJ/mol).
T is the temperature in Kelvin.

To convert the temperature from Celsius to Kelvin, we add 273.15 to the given boiling point (-34.55 °C):

T = -34.55 + 273.15 = 238.6 K

Now we can substitute the values into the formula:

ΔSvap = 19.76 kJ/mol / 238.6 K

To get the answer in J/(mol·K), we need to convert the units of ΔHvap from kJ to J by multiplying it by 1000:

ΔSvap = (19.76 kJ/mol * 1000 J/kJ) / 238.6 K

Simplifying the equation gives us the answer:

ΔSvap = 82.79 J/(mol·K)

You can't substitute 19.76 for delta Hvap and divide by -34.55 C (in kelvin, of course) to arrive at DSvap? You must want someone to punch the calculator for you.