help me please i don't understand what to do...
Calculate the molar entropy of vaporization for liquid hydrogen iodide at its boiling point -34.55 degrees celcius.
HI(l) arrows pointing to both HI(g)
Triangle H of vap= 19.76 kJ/mol
remember the equation delta G = delta H - (T x delta S)?
at the boiling point, delta G should be 0.
So you just plug given values into the equation to find delta S.
To begin, convert the temperature into Kelvin: -34.55 degrees celsius + 273 = ____K
plug into equation: 0 = 19.76 - (____K x delta S). Solve for delta S. Because the unit for delta H was kJ/mol, we need to convert the answer into from kJ/mol K into J/mol K because that's molar entropy's unit.
That should be it, I think it should be correct. Hopefully that helps a bit =).
Well, it seems like you're in a bit of a pickle! Don't worry, I'm here to help you with a touch of humor.
Calculating molar entropy of vaporization can be a real "boiling" point for many. But fear not, my friend! We'll navigate this "icy" territory together.
Firstly, take a deep breath, just like a molecule of hydrogen iodide. *Inhale... exhale...* Now, let's focus on the given information.
Given:
Temperature = -34.55 °C
∆Hvap = 19.76 kJ/mol
To calculate the molar entropy of vaporization (∆Svap), we can use the equation ∆Svap = ∆Hvap / T.
But hold on a moment! Our temperature is in Celsius, while the equation requires Kelvin. Let's convert it! Add 273.15 to -34.55 to get 238.6 K. Now we're cooking!
Now, plug in the values: ∆Svap = 19.76 kJ/mol / 238.6 K.
Calculating ∆Svap may seem overwhelming, but remember, we're just dividing two numbers here. So grab a calculator and let's do some math magic!
Once you solve it, voila! You'll have the molar entropy of vaporization for liquid hydrogen iodide at its boiling point. And just like that, we've vaporized this problem!
Remember, if things start getting too "explosive," don't hesitate to reach out for help. We're always here to add a touch of laughter to lighten the scientific load.
To calculate the molar entropy of vaporization, you need to use the following equation:
ΔSvap = ΔHvap / T
Where:
ΔSvap = molar entropy of vaporization
ΔHvap = molar enthalpy of vaporization (given as 19.76 kJ/mol)
T = temperature in Kelvin
First, convert the boiling point of hydrogen iodide from Celsius to Kelvin:
T = -34.55°C + 273.15 = 238.60 K
Now, plug the values into the equation:
ΔSvap = 19.76 kJ/mol / 238.60 K
Calculating the result:
ΔSvap = 0.0829 kJ/(mol·K)
Therefore, the molar entropy of vaporization for liquid hydrogen iodide at its boiling point of -34.55°C is approximately 0.0829 kJ/(mol·K).
Sure! I can help you with that.
To calculate the molar entropy of vaporization (ΔS_vap) for a substance at its boiling point, you need to use the equation:
ΔS_vap = ΔH_vap / T
Where ΔH_vap is the molar enthalpy of vaporization and T is the temperature in Kelvin.
In this case, you are given that the molar enthalpy of vaporization (ΔH_vap) for liquid hydrogen iodide (HI) is 19.76 kJ/mol. However, the given value is in kilojoules per mole, and we need to convert it to joules per mole for consistency.
Therefore, 19.76 kJ/mol is equal to 19.76 × 10^3 J/mol.
Now, we need to convert the temperature from Celsius to Kelvin by adding 273.15 to -34.55 degrees Celsius:
T = -34.55 + 273.15 = 238.6 K
Now that we have all the necessary values, we can calculate the molar entropy of vaporization:
ΔS_vap = (19.76 × 10^3 J/mol) / 238.6 K
By performing the calculation, you will find the molar entropy of vaporization for liquid hydrogen iodide at its boiling point.