Problem 57
Question
For a certain chemical reaction, \(\Delta H^{k}=-35.4 \mathrm{~kJ}\) and \(\Delta S^{n}=-85.5 \mathrm{~J} / \mathrm{K}\). (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?
Step-by-Step Solution
Verified Answer
(a) The reaction is exothermic since \(\Delta H^k = -35.4 \mathrm{~kJ}\) is negative. (b) The reaction leads to a decrease in the randomness or disorder of the system as \(\Delta S^n = -85.5 \mathrm{~J} / \mathrm{K}\) is negative. (c) The Gibbs free energy change at \(298 \mathrm{~K}\) is \(\Delta G^\circ = -9,961 \mathrm{~J}\). (d) The reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions as \(\Delta G^\circ\) is negative.
1Step 1: (a) Determine reaction type
Recall that an exothermic reaction emits heat and has a negative \(\Delta H^k\) value, while an endothermic reaction absorbs heat and has a positive \(\Delta H^k\) value. In this case, since the \(\Delta H^k\) value is given as \(-35.4 \mathrm{~kJ}\) (negative), the reaction is exothermic.
2Step 2: (b) Determine the change in randomness/disorder
An increase in the entropy change (\(\Delta S^n\)) corresponds to an increase in the randomness or disorder of the system, while a decrease in \(\Delta S^n\) corresponds to a decrease in the system's randomness. Since \(\Delta S^n\) is given as \(-85.5 \mathrm{~J} / \mathrm{K}\) (negative), the reaction leads to a decrease in the randomness or disorder of the system.
3Step 3: (c) Calculate Gibbs free energy change
To calculate the Gibbs free energy change (\(\Delta G^\circ\)) for the reaction at \(298 \mathrm{~K}\), we can use the following formula:
\[
\Delta G^\circ = \Delta H^k - T \Delta S^n
\]
where \(T\) is the temperature in Kelvin (K), and remember to convert \(\Delta H^k\) and \(\Delta S^n\) to the same units (Joules) before plugging the values into the formula. In this case:
\[
\Delta G^\circ = (-35.4 \times 10^3 \mathrm{~J}) - (298 \mathrm{~K} \times -85.5 \mathrm{~J} / \mathrm{K})
\]
4Step 4: Calculate Gibbs free energy using the values from above
Plugging in the values from above, we get:
\[
\Delta G^\circ = (-35.4 \times 10^3 \mathrm{~J}) - (298 \mathrm{~K} \times -85.5 \mathrm{~J} / \mathrm{K})
\]
\[
\Delta G^\circ = -35,400 \mathrm{~J} + (25,439 \mathrm{~J})
\]
\[
\Delta G^\circ = -9,961 \mathrm{~J}
\]
Thus, the Gibbs free energy change for the reaction at \(298 \mathrm{~K}\) is \(-9,961 \mathrm{~J}\).
5Step 5: (d) Determine the spontaneity of the reaction
If the Gibbs free energy change (\(\Delta G^\circ\)) of a reaction is negative, the reaction is spontaneous; if it is positive, the reaction is non-spontaneous. Since the \(\Delta G^\circ\) value we calculated above is \(-9,961 \mathrm{~J}\) (negative), the reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions.
Key Concepts
Exothermic ReactionEntropy ChangeReaction SpontaneityStandard Conditions
Exothermic Reaction
When studying chemical reactions, it's crucial to understand the concept of exothermic reactions. These reactions are characterized by the release of energy, generally in the form of heat, to their surroundings. This is noticeable when a reaction feels warm or hot to the touch. The energy involved in an exothermic reaction is quantified by a term called enthalpy change, represented by ewline ewline (ewline ewline H).
An exothermic reaction has a negative ewline ewline (ewline ewline H) value, indicating that energy is exiting the system. In other words, the total energy of the products is less than the total energy of the reactants. This is why burning wood in a fireplace, for instance, produces heat that warms the room—an example of an exothermic process.
In the given exercise, the reaction is identified as exothermic because the (H^k) value is negative (-35.4 kJ). This negative sign is the key indicator for exothermicity, and a major factor when determining the overall energy profile of a chemical reaction. Exothermic reactions, due to their energy-releasing nature, often tend to proceed spontaneously under suitable conditions.
An exothermic reaction has a negative ewline ewline (ewline ewline H) value, indicating that energy is exiting the system. In other words, the total energy of the products is less than the total energy of the reactants. This is why burning wood in a fireplace, for instance, produces heat that warms the room—an example of an exothermic process.
In the given exercise, the reaction is identified as exothermic because the (H^k) value is negative (-35.4 kJ). This negative sign is the key indicator for exothermicity, and a major factor when determining the overall energy profile of a chemical reaction. Exothermic reactions, due to their energy-releasing nature, often tend to proceed spontaneously under suitable conditions.
Entropy Change
Entropy change ( S) is a measure of the disorder or randomness in a system. It's a fundamental concept in understanding chemical processes, as it provides insight into the degree of chaos at the particle level. When a substance moves from a more ordered state to a less ordered one, like a solid melting into a liquid, its entropy increases.
In chemistry, the sign of the ( S^n) value is critical: a positive value indicates an increase in disorder, while a negative value signals a decrease. In the context of chemical reactions, breaking bonds, mixing of gases, or dissolving a solute into a solvent typically lead to higher entropy.
In the exercise, the negative entropy change (-85.5 J/K) tells us that the system becomes more ordered as the reaction progresses. This could involve formation of a solid from liquids or gases, or the reactant molecules arranging into a more structured form in the products.
In chemistry, the sign of the ( S^n) value is critical: a positive value indicates an increase in disorder, while a negative value signals a decrease. In the context of chemical reactions, breaking bonds, mixing of gases, or dissolving a solute into a solvent typically lead to higher entropy.
In the exercise, the negative entropy change (-85.5 J/K) tells us that the system becomes more ordered as the reaction progresses. This could involve formation of a solid from liquids or gases, or the reactant molecules arranging into a more structured form in the products.
Reaction Spontaneity
Delving into reaction spontaneity, it’s all about determining whether a reaction will proceed without the need for external energy. This concept can be somewhat counterintuitive as it does not necessarily relate to the speed of the reaction—spontaneous reactions can be fast or slow.
To figure out if a reaction is spontaneous, we look at Gibbs free energy change ( G^), which encompasses both the reaction's enthalpy ( H^k) and its entropy ( S^n). The temperature ( T) of the system also plays a vital role. The equation ewline ewline G^= H^k - T S^n) helps us calculate G^. For a process at constant temperature and pressure, a negative G^ indicates that the reaction is spontaneous, and a positive value suggests it is non-spontaneous.
In the given solution, the calculated Gibbs free energy change (-9,961 J) is negative, which implies that the reaction is indeed spontaneous at 298 K. It's pivotal to remember that a negative G^ does not say how fast the reaction occurs, just that it is thermodynamically favorable and can occur without external intervention under the given conditions.
To figure out if a reaction is spontaneous, we look at Gibbs free energy change ( G^), which encompasses both the reaction's enthalpy ( H^k) and its entropy ( S^n). The temperature ( T) of the system also plays a vital role. The equation ewline ewline G^= H^k - T S^n) helps us calculate G^. For a process at constant temperature and pressure, a negative G^ indicates that the reaction is spontaneous, and a positive value suggests it is non-spontaneous.
In the given solution, the calculated Gibbs free energy change (-9,961 J) is negative, which implies that the reaction is indeed spontaneous at 298 K. It's pivotal to remember that a negative G^ does not say how fast the reaction occurs, just that it is thermodynamically favorable and can occur without external intervention under the given conditions.
Standard Conditions
The term 'standard conditions' in chemistry refers to a set of predefined reference conditions that allow for comparison and prediction of chemical behavior. Standard conditions usually consist of a pressure of 1 atmosphere and a specified temperature, often 25^C (298 K.)
Under these conditions, the properties of substances, like Gibbs free energy change ( G^), can be estimated with standard tables. This ensures consistency across laboratories and textbooks when measuring or calculating thermodynamic quantities.
The exercise determines the spontaneity of the reaction specifically at 298 K, signifying that we are considering standard temperature. This casts light on how environmental factors can dramatically influence the spontaneity and behavior of chemical reactions. It is also why it’s expressly pinpointed that the reaction's spontaneity is ascertained ‘under standard conditions’, confirming the importance of context when discussing thermodynamic properties in chemistry.
Under these conditions, the properties of substances, like Gibbs free energy change ( G^), can be estimated with standard tables. This ensures consistency across laboratories and textbooks when measuring or calculating thermodynamic quantities.
The exercise determines the spontaneity of the reaction specifically at 298 K, signifying that we are considering standard temperature. This casts light on how environmental factors can dramatically influence the spontaneity and behavior of chemical reactions. It is also why it’s expressly pinpointed that the reaction's spontaneity is ascertained ‘under standard conditions’, confirming the importance of context when discussing thermodynamic properties in chemistry.
Other exercises in this chapter
Problem 55
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View solution Problem 61
Using data from Appendix \(C\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at \(298 \mathrm{~K}\)
View solution