Find out whether it is possible to reduce \(\mathrm{MgO}\) using carbon at \(298 \mathrm{~K}\). If not, at what temperature it becomes spontaneous. For reaction, \(\mathrm{MgO}(s)+\mathrm{C}(s) \longrightarrow \mathrm{Mg}(s)+\mathrm{CO}(g), \Delta H^{\circ}=+491.18 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\Delta S^{0}=197.67 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Short Answer

Expert verified
Calculated \(\Delta G\) at \(298 \mathrm{K}\) comes out positive, indicating the process is non-spontaneous at room temperature. The reaction becomes spontaneous at a temperature higher than the one calculated in Step 5.

Step by step solution

01

Understanding the Gibbs Free Energy

Gibbs Free Energy, denoted as \(\Delta G\), is used to determine whether a process is spontaneous. The equation for Gibbs Free Energy is \[\Delta G = \Delta H - T\Delta S\] where \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy. If \(\Delta G < 0\), the process is spontaneous; if \(\Delta G > 0\), it is non-spontaneous.
02

Calculating Gibb's Free Energy at 298 K

First, we will calculate the value of \(\Delta G\) at \(298 \mathrm{K}\) using the given values for \(\Delta H\) and \(\Delta S\). Plug the values into the equation: \[\Delta G = +491.18 \times 10^3 \mathrm{J/mol} - (298 \mathrm{K})(197.67 \mathrm{J/K/mol})\].
03

Evaluate the spontaneity at 298 K

Calculate \(\Delta G\) using the values from Step 2: \[\Delta G = 491180 \mathrm{J/mol} - (298)(197.67 \mathrm{J/K/mol})\] and assess whether \(\Delta G\) is positive or negative.
04

Find the Temperature for Spontaneity

If the process is non-spontaneous at \(298 \mathrm{K}\), set \(\Delta G\) to zero and solve for \(T\) to find the temperature at which the process becomes spontaneous: \[0 = \Delta H - T\Delta S\]. Rearrange the equation to solve for the temperature: \[T = \frac{\Delta H}{\Delta S}\].
05

Calculating the Temperature

Calculate the temperature using the values for \(\Delta H\) and \(\Delta S\): \[T = \frac{491180 \mathrm{J/mol}}{197.67 \mathrm{J/K/mol}}\].
06

Conclusion

Gather the results from Steps 3 and 5, evaluating the spontaneity at \(298 \mathrm{K}\) and determining the temperature at which the reaction becomes spontaneous if required.

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Key Concepts

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

Spontaneity of Reactions
Understanding the spontaneity of chemical reactions is crucial in predicting whether a reaction will occur under a set of conditions. A reaction is spontaneous if it occurs without being driven by some external force. The Gibbs Free Energy (abla G) formula provides insight into this spontaneity. For a reaction at constant temperature and pressure, if abla G is less than zero (abla G < 0), the process is considered spontaneous. On the other hand, a positive abla G (> 0) indicates a non-spontaneous reaction, implying it needs external energy to proceed. At equilibrium, abla G equals zero, and no net reaction occurs; the forward and reverse reactions occur at the same rate.

The exercise aimed at determining the spontaneity of reducing magnesium oxide (MgO) with carbon. The calculation of Gibbs Free Energy at standard conditions (298 K) with the given enthalpy (H^{0}) and entropy (S^{0}) values allowed us to assess whether the reaction is spontaneous at this temperature. By resolving the Gibbs formula with abla G = abla H - Tabla S, we can infer the spontaneity based on the sign of the resultant abla G.
Entropy and Enthalpy
In thermodynamics, entropy (abla S) and enthalpy (abla H) are fundamental concepts that explain the energy changes in chemical reactions. Entropy relates to the degree of disorder or randomness in a system; it is a measure of the number of specific ways a system can be arranged. Generally, natural processes tend to increase the entropy of the universe.

Enthalpy, on the other hand, is a measure of the total energy of a thermodynamic system, including both internal energy and the energy required to displace its environment to make room for its volume. A positive change in enthalpy (abla H > 0) indicates that a reaction is endothermic, absorbing heat from its surroundings. Conversely, a negative abla H (< 0) signifies an exothermic reaction, where energy is released as heat to the surroundings.

The given exercise provides the values of both abla H and abla S for the reaction between magnesium oxide and carbon. By considering these values within the Gibbs Free Energy equation, we were able to calculate how they influence the reaction's spontaneity.
Chemical Equilibrium
Chemical equilibrium is a state in a reversible chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of reactants and products over time. At equilibrium, the Gibbs Free Energy of the system is at its minimum, and thus the free energy change (abla G) for the reaction is zero.

The magnitude of the equilibrium constant (K_{eq}) provides insight into the position of equilibrium. A large K_{eq} value indicates a reaction that favors the formation of products, while a small K_{eq} value suggests a reaction favoring reactants. In the context of our exercise, by setting the Gibbs Free Energy equation to zero and solving for the temperature (T), we effectively find the temperature at which the system would reach equilibrium for the given reaction if it could achieve equilibrium at all.
Thermodynamics in Physical Chemistry
The principles of thermodynamics are the foundation of physical chemistry, explaining the energy changes and exchanges that occur during chemical reactions and phase transitions. Among these, the first and second laws of thermodynamics govern the concepts of energy conservation and entropy, respectively.

In our exercise, thermodynamics helps predict the conditions under which the reduction of MgO by carbon would spontaneously occur. By applying the equations and concepts of thermodynamics, such as the calculation of Gibbs Free Energy, we determined not only the spontaneous nature of the reaction at a given temperature but also the theoretical temperature at which the reaction would become spontaneous, based on the enthalpy and entropy values provided.

The exercise guides students to leverage these thermodynamics concepts to assess reaction spontaneity and equilibrium, enhancing their understanding of how energy and matter interact in chemical processes.

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Most popular questions from this chapter

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 \(35^{\circ} \mathrm{C}\).

Calculate the equilibrium constant \(K_{\mathrm{p}}\) for the reaction given below if \(\Delta G^{\circ}=-10.632 \mathrm{~kJ}\) at \(300 \mathrm{~K}\) $$ \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} $$

Ethanol boils at \(78.4^{\circ} \mathrm{C}\) and the enthalpy of vaporisation of ethanol is \(42.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Calculate the entropy of vaporisation of ethanol.

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 \(\Delta H^{\circ}=28.40 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

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