What will be the melting point of \(\mathrm{KCl}\) if enthalpy change for the reaction is \(7.25 \mathrm{~J} \mathrm{~mol}^{-1}\) and entropy change is \(0.007 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) ? (a) \(1835.2 \mathrm{~K}\) (b) \(173 \mathrm{~K}\) (c) \(1035.7 \mathrm{~K}\) (d) \(1285.2 \mathrm{~K}\)

Short Answer

Expert verified
The melting point of KCl is (c) 1035.7 K.

Step by step solution

01

Understanding the Concept

The melting point of a substance can be found using the equation that relates the enthalpy (ΔH) and entropy (ΔS) changes during the phase transition at a temperature (T), known as the Gibbs free energy (ΔG). At the melting point, the Gibbs free energy of the phase transition is zero (ΔG = 0), so the equation ΔG = ΔH − TΔS can be rearranged to solve for the melting point temperature T.
02

Rearrange the Gibbs Free Energy Equation to Solve for Temperature

Rearrange the Gibbs free energy equation ΔG = ΔH − TΔS to solve for the temperature (T) at the melting point, where ΔG = 0. The rearranged equation is T = ΔH / ΔS.
03

Calculate the Melting Point

Insert the given enthalpy change (ΔH = 7.25 J/mol) and entropy change (ΔS = 0.007 J/K·mol) into the equation and solve for the temperature. T = (7.25 J/mol) / (0.007 J/K·mol) = 1035.7 K.

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy, symbolized as \(\Delta G\), is a thermodynamic quantity that provides invaluable insights into the spontaneity of chemical reactions and phase transitions, such as melting or freezing. When calculating melting points, \(\Delta G\) becomes particularly significant because it must be zero at equilibrium, where the solid and liquid phases coexist. A zero value indicates that there is no net energy advantage for the substance to exist in either the solid or liquid state; they are in perfect balance.

Since \(\Delta G\) is defined by the equation \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) is the enthalpy change, \(T\) is the temperature, and \(\Delta S\) is the entropy change, understanding this relationship is crucial in melting point analysis. At the melting point, since we set \(\Delta G = 0\), the temperature \(T\) can be directly calculated by rearranging the equation to \(T = \Delta H / \Delta S\), thus seamlessly linking all these thermodynamic concepts.
Enthalpy Change
Enthalpy change, denoted as \(\Delta H\), represents the heat exchanged during a chemical reaction or phase transition at constant pressure. This value is a core factor in determining a substance's melting point. True to its definition, it can be either positive or negative; a positive \(\Delta H\) implies that the system absorbs heat (endothermic process), whereas a negative \(\Delta H\) indicates heat release (exothermic process).

During melting, a solid absorbs heat as it transitions to a liquid state, which is why we consider enthalpy change for calculating the melting point. In the case of \(\mathrm{KCl}\), the enthalpy change during melting is 7.25 joules per mole, indicating the amount of energy absorbed when one mole of \(\mathrm{KCl}\) melts at a constant pressure.
Entropy Change
Entropy change, symbolized as \(\Delta S\), is a measure of the disorder or randomness within a thermodynamic system. It's an essential concept in thermodynamics as it influences the feasibility and direction of a chemical process, including phase transitions like melting. In the context of melting point calculation, \(\Delta S\) represents the change in disorder occurring as a solid turns into a liquid – generally an increase due to the greater freedom of movement among liquid particles.

An important point to remember is that a higher \(\Delta S\) value means a bigger increase in disorder. For the melting of \(\mathrm{KCl}\), the entropy change is 0.007 joules per kelvin per mole. This quantifies the extent to which disorder increases when \(\mathrm{KCl}\) melts at the molecular level.

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Most popular questions from this chapter

The amount of heat evolved when \(0.50 \mathrm{~m}_{0} \mathrm{e}\) of \(\mathrm{HCl}\) is mixed with \(0.30\) mole of \(\mathrm{NaOH}\) solution is (a) \(57.1 \mathrm{~kJ}\) (b) \(28.55 \mathrm{~kJ}\) (c) \(11.42 \mathrm{~kJ}\) (d) \(17.13 \mathrm{~kJ}\)

What will be \(\Delta H\) for the reaction, \(\mathrm{CH}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{C}+2 \mathrm{H}+2 \mathrm{Cl} ?\) \(\left(B \cdot E\right.\). of \(\mathrm{C}-\mathrm{H}\) and \(\mathrm{C}-\mathrm{Cl}\) bonds are \(416 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(325 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively) (a) \(832 \mathrm{~kJ}\) (b) \(1482 \mathrm{~kJ}\) (c) \(650 \mathrm{~kJ}\) (d) \(1855 \mathrm{~kJ}\)

200 joules of heat was supplied to a system at constant volume. It resulted in the increase in temperature of the system from 298 to \(323 \mathrm{~K}\). What is the change in internal energy of the system? (a) \(400 \mathrm{~J}\) (b) \(200 \mathrm{~J}\) (c) \(50 \mathrm{~J}\) (d) \(150 \mathrm{~J}\)

The equilibrium constant for a reaction is \(10 .\) What will be the value of \(\Delta G^{\circ}\) at \(300 \mathrm{~K} ?\) (a) \(-5.74 \mathrm{~kJ}\) (b) \(-574 \mathrm{~kJ}\) (c) \(+11.48 \mathrm{~kJ}\) (d) \(+5.74 \mathrm{~kJ}\)

At \(373 \mathrm{~K}\), steam and water are in equilibrium and \(\Delta H=40.98 \mathrm{~kJ} \mathrm{~mol}^{-1}\). What will be \(\Delta S\) for conversion of water into steam? \(\mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow \mathrm{H}_{2} \mathrm{O}_{(g)}\) (a) \(109.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(31^{-1} \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(21.93 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (d) \(326 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

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