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Chapter 18: Problem 60

The molar heat of vaporization of ethanol is \(39.3 \mathrm{~kJ} / \mathrm{mol}\) and the boiling point of ethanol is \(78.3^{\circ} \mathrm{C}\) Calculate \(\Delta S\) for the vaporization of \(0.50 \mathrm{~mol}\) ethanol.

Short Answer

Expert verified
The entropy change for the vaporization of 0.50 mol ethanol is 111.82 J/mol.K.

Step by step solution

01

Convert temperature to Kelvin

To convert °C to Kelvin, you add 273.15 to the °C value. For us, this means \(T = 78.3°C + 273.15 = 351.45K\).
02

Convert heat of vaporization to J/mol

Given value of heat of vaporization is in kJ/mol. To do calculations correctly, we need to make sure all the units match. To convert kJ to J, multiply the kJ value by 1000. Thus, \(\Delta H_vap = 39.3 kJ/mol × 1000 = 39300 J/mol\).
03

Calculate entropy change using the formula

Now we can substitute the values into the formula \(ΔS = \(\frac{q_{rev}}{T}\)\). This gives: \(\Delta S = \(\frac{39300 J/mol}{351.45K}\) = 111.82 J/mol.K\).

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Most popular questions from this chapter

The reaction \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) proceeds spontaneously at \(25^{\circ} \mathrm{C}\) even though there is a decrease in the number of microstates of the system (gases are converted to a solid). Explain.

Arrange the following substances ( 1 mole each) in order of increasing entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Ne}(g)\), (b) \(\mathrm{SO}_{2}(g),\) (c) \(\mathrm{Na}(s)\) (d) \(\mathrm{NaCl}(s)\) (e) \(\mathrm{H}_{2}(g)\). Give the reasons for your arrangement.

A student placed \(1 \mathrm{~g}\) of each of three compounds \(\mathrm{A}\) \(\mathrm{B},\) and \(\mathrm{C}\) in a container and found that after 1 week no change had occurred. Offer some possible explanations for the fact that no reactions took place. Assume that \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are totally miscible liquids.

Entropy has sometimes been described as "time's arrow" because it is the property that determines the forward direction of time. Explain.

(a) Calculate \(\Delta G^{\circ}\) and \(K_{P}\) for the following equilibrium reaction at \(25^{\circ} \mathrm{C}\). The \(\Delta G_{f}^{\circ}\) values are 0 for \(\mathrm{Cl}_{2}(g),-286 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{3}(g),\) and \(-325 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{5}(g)\) $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$. (b) Calculate \(\Delta G\) for the reaction if the partial pressures of the initial mixture are \(P_{\mathrm{PCl}_{5}}=0.0029 \mathrm{~atm}\) \(P_{\mathrm{PCl}_{3}}=0.27 \mathrm{~atm},\) and \(P_{\mathrm{Cl}_{2}}=0.40 \mathrm{~atm}\).
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