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If a reaction can be carried out only by electrolysis, which of the following changes in a thermodynamic property must apply: (a) \(\Delta H>0 ;\) (b) \(\Delta S>0\) (c) \(\Delta G=\Delta H ;\) (d) \(\Delta G>0 ?\) Explain.

Short Answer

Expert verified
(d) For a reaction to only be carried out by electrolysis, the change in Gibbs free energy (\(\Delta G\)) needs to be greater than zero (\(\Delta G > 0\)).

Step by step solution

01

Understanding the Thermodynamic Properties

Before addressing the specific options in the exercise, it's essential to understand what each sign of these properties means: \n- \(\Delta H\) represents the change in enthalpy (the heat content) of a reaction. If \(\Delta H > 0\), heat is absorbed (endothermic process) and if \(\Delta H < 0\), heat is released (exothermic process). - \(\Delta S\) stands for the change in entropy (the disorder of the system). If \(\Delta S > 0\), the system is becoming more disordered. If \(\Delta S < 0\), the system is becoming more ordered.- \(\Delta G\) denotes the change in Gibbs free energy, which is the maximum reversible work done at constant temperature and pressure. It's calculated as \(\Delta G = \Delta H - T\Delta S\). If \(\Delta G < 0\), the reaction is spontaneous. If \(\Delta G > 0\), the reaction is non-spontaneous and requires external energy.
02

Understanding Electrolysis

Electrolysis involves the use of an electric current to drive a non-spontaneous chemical reaction - a reaction that otherwise wouldn’t occur without the addition of energy. So in terms of thermodynamics, for a reaction to only be feasible via electrolysis, it must be non-spontaneous, i.e., \(\Delta G > 0\).
03

Evaluating the Options

With the above insights, it can be seen that the relevant answer is (d) \(\Delta G > 0\). Neither a positive \(\Delta H\), nor a positive \(\Delta S\), nor \(\Delta G = \Delta H\) would necessarily imply that the reaction could only occur via electrolysis.

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Key Concepts

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

Enthalpy
In the realm of thermodynamics, enthalpy (\(\Delta H\) ) serves as a measure of the total heat content of a system. Imagine it as a form of energy storage capacity for heat in a chemical reaction. **Key Points to Understand Enthalpy:**
  • When \(\Delta H > 0\), the reaction requires energy absorption, and is known as an endothermic process. This means the system takes in heat from its surroundings.
  • Conversely, when \(\Delta H < 0\) , it's an exothermic process. This indicates the system releases heat to its surroundings.
Knowing the enthalpy change helps predict whether a reaction will absorb or release heat, which is crucial for understanding reaction energies.The change in enthalpy does not always correlate with the feasibility of a reaction occurring without external energy, but it does provide insights into the energy dynamics involved.
Entropy
Entropy (\(\Delta S\)) is a fascinating thermodynamic property because it quantifies the amount of disorder or randomness in a system.**How to Grasp Entropy:**
  • An increase in entropy ( \(\Delta S > 0\) ) suggests that a system is becoming more disordered.
  • When \(\Delta S < 0\), the system becomes more structured and orderly.
Consider a melting ice cube: as it transitions from a solid to a liquid, it becomes more disordered, increasing entropy.Entropy plays a pivotal role in spontaneous reactions, interacting with both enthalpy and temperature to determine the overall energy balance. Despite influencing spontaneity, entropy itself does not dictate whether a reaction will occur without external energy. Instead, it contributes to the calculation of Gibbs free energy, which we'll explore next.
Gibbs Free Energy
Gibbs free energy (\(\Delta G\)) is derived from both enthalpy and entropy. It provides an overarching view of the feasibility and spontaneity of a reaction at constant temperature and pressure.**Understanding Gibbs Free Energy:**
  • The relationship is given by the equation \(\Delta G = \Delta H - T\Delta S\) , where \(T\) is the temperature in Kelvin.
  • A negative Gibbs free energy ( \(\Delta G < 0\) ) indicates a spontaneous reaction. This implies the reaction can occur naturally without external input.
  • If \(\Delta G > 0\) , the reaction is non-spontaneous. This means it requires additional energy, like in electrolysis, to proceed.
Gibbs free energy is pivotal in determining if a reaction is possible without energy input. It also integrates enthalpy and entropy, offering a comprehensive answer to whether environmental conditions favor a reaction's progress. Understanding \(\Delta G\) helps chemists and physicists gauge the direction and capability of chemical processes, essential for innovations and applications across scientific fields.

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Most popular questions from this chapter

For one of the following reactions, \(K_{c} K_{p}=K .\) Identify that reaction. For the other two reactions, what is the relationship between \(K_{c}, \bar{K}_{\mathrm{p}},\) and \(K ?\) Explain. (a) \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\) (b) \(\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g})\) (c) \(\mathrm{NH}_{4} \mathrm{HCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1)\)

Calculate the equilibrium constant and Gibbs energy for the reaction \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) at \(483 \mathrm{K}\) by using the data tables from Appendix D. Are the values determined here different from or the same as those in exercise \(35 ?\) Explain.

The following data are given for the two solid forms of \(\mathrm{HgI}_{2}\) at \(298 \mathrm{K}\). $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ} & \Delta G_{f,}^{\circ} & S^{\circ} \\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{HgI}_{2} \text { (red) } & -105.4 & -101.7 & 180 \\ \mathrm{Hg} \mathrm{I}_{2} \text { (yellow) } & -102.9 & (?) & (?) \\ \hline \end{array}$$ Estimate values for the two missing entries. To do this, assume that for the transition \(\mathrm{HgI}_{2}(\mathrm{red}) \longrightarrow\) \(\mathrm{HgI}_{2}(\text { yellow }),\) the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(25^{\circ} \mathrm{C}\) have the same values that they do at the equilibrium temperature of \(127^{\circ} \mathrm{C}\).

For the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\) \(K_{\mathrm{c}}=2.8 \times 10^{2}\) at \(1000 \mathrm{K}\) (a) What is \(\Delta G^{\circ}\) at \(1000 \mathrm{K} ?\left[\text { Hint: What is } \mathrm{K}_{\mathrm{p}} ?\right]\) (b) If \(0.40 \mathrm{mol} \mathrm{SO}_{2}, 0.18 \mathrm{mol} \mathrm{O}_{2},\) and \(0.72 \mathrm{mol} \mathrm{SO}_{3}\) are mixed in a 2.50 L flask at \(1000 \mathrm{K}\), in what direction will a net reaction occur?

The Gibbs energy change of a reaction can be used to assess (a) how much heat is absorbed from the surroundings; (b) how much work the system does on the surroundings; (c) the net direction in which the reaction occurs to reach equilibrium; (d) the proportion of the heat evolved in an exothermic reaction that can be converted to various forms of work.

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