Asked by trista
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
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
Answers
Answered by
Anonymous
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 =).
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 =).
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