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Chapter 11: Problem 78

The molar heats of fusion and sublimation of molecular iodine are \(15.27 \mathrm{~kJ} / \mathrm{mol}\) and \(62.30 \mathrm{~kJ} / \mathrm{mol}\), respectively. Estimate the molar heat of vaporization of liquid iodine.

Short Answer

Expert verified
The molar heat of vaporization of iodine is approximately \(47.03 \, \mathrm{kJ/mol}\).

Step by step solution

01

Understand Hess's Law

Hess's Law of Constant Heat Summation states when a reaction can be expressed as the sum of a series of steps, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual steps.
02

Apply Hess's Law to the problem

The problem can be visualized as: \nSolid iodine melts (fusion) into liquid iodine and then boils (vaporizes) into gaseous iodine. Alternatively, solid iodine can directly sublimate into gaseous iodine.\nThe molar heat of fusion is the heat change when one mole of solid iodine melts into liquid iodine, and the molar heat of sublimation is the heat change when one mole of solid iodine transforms directly into gaseous iodine.\nThe molar heat of vaporization (which we aim to calculate) is the heat change when one mole of liquid iodine vaporizes into gaseous iodine.
03

Calculate the molar heat of vaporisation

According to Hess's Law, the molar heat of sublimation is equal to the sum of the molar heat of fusion and vaporization: \n\n\[\Delta H_{sublimation}= \Delta H_{fusion} + \Delta H_{vaporization}\]\n\nWe can rearrange the equation to calculate the molar heat of vaporization as follows:\n\n\[\Delta H_{vaporization}= \Delta H_{sublimation} − \Delta H_{fusion}\]\n\nThen substituting the given values:\n\n\[\Delta H_{vaporization} = 62.30 kJ/mol - 15.27 kJ/mol = 47.03 kJ/mol\]

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