The enthalpy of vaporisation of liquid diethyl ether \(-\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\), is \(26.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at its boiling point \(\left(35.0^{\circ} \mathrm{C}\right)\). Calculate \(\Delta S\) for conversion of : (a) liquid to vapour, and \(\quad\) (b) vapour to liquid at \(35^{\circ} \mathrm{C}\).

Short Answer

Expert verified
\(\Delta S\) for liquid to vapour is +84.3 J/(mol\cdot K) and for vapour to liquid is -84.3 J/(mol\cdot K).

Step by step solution

01

Identify the Given Information

To solve the problem, recognize that the enthalpy of vaporization \(\Delta H_{vap}\) is given as 26.0 kJ/mol at the boiling point of 35.0°C. The formula to calculate the entropy change \(\Delta S\) for the phase transition is defined as \(\Delta S = \frac{\Delta H_{vap}}{T}\), where T is the temperature in Kelvin.
02

Convert the Temperature to Kelvin

The temperature must be in Kelvin for the entropy calculation. To convert from Celsius to Kelvin, use the formula \(T(K) = T(°C) + 273.15\). Thus, \(T = 35.0°C + 273.15 = 308.15 K\).
03

Calculate the Change in Entropy for Liquid to Vapour

Now, calculate the change in entropy for the conversion of liquid to vapour using the formula \(\Delta S = \frac{\Delta H_{vap}}{T}\). Substituting the values, we get \(\Delta S = \frac{26.0 kJ/mol}{308.15 K}\).
04

Express the Entropy Change in Correct Units

The enthalpy of vaporization is given in kJ/mol, but entropy is typically expressed in J/(mol·K). To convert kJ to J, multiply by 1000: \(\Delta S = \frac{26.0 \times 1000 J/mol}{308.15 K}\).
05

Solve for the Entropy Change for Liquid to Vapour

Now perform the division to find \(\Delta S\): \(\Delta S = \frac{26000 J/mol}{308.15 K} = 84.3 J/(mol\cdot K)\). This is the entropy change for liquid to vapour.
06

Calculate the Change in Entropy for Vapour to Liquid

To find the entropy change for the conversion of vapour to liquid, simply change the sign of the \(\Delta S\) obtained for liquid to vapour, because the process is exactly the opposite: \(\Delta S_{vapour\ to\ liquid} = -\Delta S_{liquid\ to\ vapour} = -84.3 J/(mol\cdot K)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy Change Calculation
Understanding entropy change calculation is central when studying thermodynamics. Entropy, denoted as \( \Delta S \), is a measure of disorder or randomness within a system. During a phase transition, such as the vaporization of a liquid, the system's entropy changes.

Using the formula \( \Delta S = \frac{\Delta H_{vap}}{T} \), where \( \Delta H_{vap} \) is the enthalpy of vaporization and T is the temperature in Kelvin, you can calculate the change in entropy. It's important to note that this equation assumes a reversible process and that the temperature remains constant during the change of phase.\
\
\In the case of liquid diethyl ether vaporizing, the positive value of \( \Delta S \) indicates an increase in entropy. This is due to the molecules in the vapor phase being much more disordered than in the liquid phase. Conversely, when a vapor condenses to a liquid, the entropy decreases, signifying a shift towards order, which is reflected by a negative entropy change.
Phase Transition
Phase transitions are transformations between different states of matter: solids to liquids (melting), liquids to gases (vaporization), and vice versa for freezing and condensation. Each transition involves a change in energy, specifically the enthalpy of the substance, and a change in entropy, as the orderliness of the system's particles changes.

For instance, when diethyl ether vaporizes (liquid to gas), it requires energy which is absorbed from the surroundings, corresponding to the enthalpy of vaporization \( \Delta H_{vap} \). This process increases the distance between particles, resulting in more possible arrangements (states), and hence a higher entropy \( \Delta S \) of the system. Remember, every phase transition has an associated \( \Delta S \) value, and signifying the direction of the entropy change helps in predicting the nature of the process—whether it is endothermic or exothermic.
Temperature Conversion to Kelvin
Temperature is a key variable in thermodynamic equations, and it's vital to use the absolute temperature scale, which is Kelvin (K), to ensure accuracy in calculations. The formula for converting Celsius to Kelvin is straightforward: \( T(K) = T(°C) + 273.15 \).

Why is this important? The Kelvin scale starts at absolute zero, the theoretical point where particles have minimum thermal motion, making it the natural choice in scientific research. This scale allows thermodynamic equations, like those used to calculate the entropy change for phase transitions, to work universally, without negative temperature values that would be nonsensical in those contexts.

For diethyl ether's boiling point at 35.0°C, converting to Kelvin gives us 308.15 K, which is then used in our entropy change calculation. This conversion forms the base for accurate and meaningful interpretations of thermal phenomena.

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Most popular questions from this chapter

Calculate the entropy change for the conversion of following: (a) \(1 \mathrm{~g}\) ice to water at \(273 \mathrm{~K}, \Delta H_{\mathrm{f}}\) for ice \(=6.025 \mathrm{~kJ} \mathrm{~mol}^{-1}\). (b) \(36 \mathrm{~g}\) water to vapour at \(373 \mathrm{~K} ; \Delta H_{\mathrm{v}}\) for \(\mathrm{H}_{2} \mathrm{O}=40.63 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

The equilibrium constant for the reaction given below is \(2.0 \times 10^{-7}\) at \(300 \mathrm{~K}\). Calculate the standard free energy change for the reaction; $$ \mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} $$ Also, calculate the standard entropy change if \(\Delta H^{\circ}=28.40 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

Calculate the change of entropy, \(\Delta_{r} S^{\circ}\) at \(298 \mathrm{~K}\) for the reaction in which urea is formed from \(\mathrm{NH}_{3}\) and \(\mathrm{CO}_{2}\). \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) .\) The standard entropy of \(\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q), \mathrm{CO}_{2}(g), \mathrm{NH}_{3}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are \(174.0\), \(213.7,192.3\) and \(69.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively.

During \(200 \mathrm{~J}\) work done on the system, \(140 \mathrm{~J}\) of heat is given out. Calculate the change in internal energy.

$$ \begin{aligned} &\text { Calculate the value of } \Delta G \text { at } 700 \mathrm{~K} \text { for the reaction } n X \longrightarrow m B .^{-1}\\\ &\text { Given that value of } \Delta H=-113 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { and } \Delta S=-145 \mathrm{JK} \mathrm{mol}^{-1} \text { . } \end{aligned} $$

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