The molar heats of sublimation and fusion of iodine are 62.3kjmol and 15.3kjmol respectively calculate the molar heat of vaporization of liquid iodide?
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Sublimation is going directly from solid to gas. Hess' Law says that no matter which pathway you take to get from point A to point B (in this case, from solid to gas) that the delta H (Hprod-Hrxt) will be the same. SO.
H(solid to gas) = H(solid to liquid) + H(liquid to gas)
Hsublimation = Hfusion + Hvaporization. 62.3 = 15.3 + Hvap
THerefore Hvap = 62.3 - 15.3 = 47 kJ/mol
there should be little triangles (deltas) in front of all the H's listed above.
H(solid to gas) = H(solid to liquid) + H(liquid to gas)
Hsublimation = Hfusion + Hvaporization. 62.3 = 15.3 + Hvap
THerefore Hvap = 62.3 - 15.3 = 47 kJ/mol
there should be little triangles (deltas) in front of all the H's listed above.
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More informatiOn required.