Problem 51
Question
51\. Determine whether the system \(\Delta H_{\text { system }}=-20.5 \mathrm{kJ}, T=298 \mathrm{K},\) and \(\Delta S_{\text { sytem }}=-35.0 \mathrm{J} / \mathrm{K}\) is spontaneous or nonspontaneous.
Step-by-Step Solution
Verified Answer
The system is spontaneous because \( \Delta G = -10.07 \text{ kJ} \).
1Step 1: Use Gibbs Free Energy Equation
To determine if a process is spontaneous, we calculate the change in Gibbs free energy, \( \Delta G \). The equation to use 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.
2Step 2: Convert Units for Entropy
Convert \( \Delta S \) from \( \mathrm{J/K} \) to \( \mathrm{kJ/K} \) because the enthalpy \( \Delta H \) is in kilojoules. Thus, \( -35.0 \mathrm{J/K} = -0.035 \mathrm{kJ/K} \).
3Step 3: Substitute Values into the Gibbs Equation
Now, substitute the values into the Gibbs equation: \[\Delta G = (-20.5 \text{ kJ}) - (298 \text{ K}) \times (-0.035 \text{ kJ/K})\]
4Step 4: Calculate the Gibbs Free Energy Change
Perform the multiplication and subtraction: \[\Delta G = -20.5 \text{ kJ} + 10.43 \text{ kJ} = -10.07 \text{ kJ}\]
5Step 5: Determine Spontaneity
A negative \( \Delta G \) indicates that the process is spontaneous. Since \( \Delta G = -10.07 \text{ kJ} \), the system is spontaneous.
Key Concepts
Enthalpy ChangeEntropy ChangeSpontaneity in Chemical Reactions
Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), represents the total energy change in a system during a chemical reaction. It's measured in kilojoules (kJ) per mole or sometimes given in joules (J) depending on the context.
In any chemical process, energy is either absorbed or released:
Such a release of energy tends to favor spontaneous reactions, but other factors, like entropy and temperature, also play a significant role in determining spontaneity.
In any chemical process, energy is either absorbed or released:
- If \( \Delta H \) is negative, the reaction is exothermic, releasing energy to the surroundings.
- If \( \Delta H \) is positive, the reaction is endothermic, absorbing energy from the surroundings.
Such a release of energy tends to favor spontaneous reactions, but other factors, like entropy and temperature, also play a significant role in determining spontaneity.
Entropy Change
Entropy, symbolized as \( \Delta S \), is a measure of disorder or randomness in a system. Generally expressed in joules per Kelvin (J/K) or kilojoules per Kelvin (kJ/K), it provides insight into the distribution of energy within a system.
When evaluating reactions:
However, both changes in entropy and enthalpy must be considered with temperature to fully determine a process's spontaneity.
When evaluating reactions:
- If \( \Delta S \) is positive, the disorder in the system increases, potentially favoring spontaneity.
- If \( \Delta S \) is negative, the system becomes less disordered, which can sometimes oppose spontaneity.
However, both changes in entropy and enthalpy must be considered with temperature to fully determine a process's spontaneity.
Spontaneity in Chemical Reactions
Spontaneity in chemical reactions depends on the Gibbs Free Energy change, \( \Delta G \). A key formula for determining spontaneity is:\[ \Delta G = \Delta H - T \Delta S \]Where:
In the exercise above, substituting the values into the formula gives:\[ \Delta G = (-20.5 \text{ kJ}) - (298 \text{ K}) \times (-0.035 \text{ kJ/K}) = -10.07 \text{ kJ} \]A result of \(-10.07 \text{ kJ}\) reveals a spontaneous process at the specified temperature.
This example illustrates how enthalpy, entropy, and temperature collectively influence whether a chemical reaction proceeds on its own.
- \(\Delta G\) is the Gibbs Free Energy change.
- \(\Delta H\) is the change in enthalpy.
- \(T\) is the temperature in Kelvin.
- \(\Delta S\) is the change in entropy.
In the exercise above, substituting the values into the formula gives:\[ \Delta G = (-20.5 \text{ kJ}) - (298 \text{ K}) \times (-0.035 \text{ kJ/K}) = -10.07 \text{ kJ} \]A result of \(-10.07 \text{ kJ}\) reveals a spontaneous process at the specified temperature.
This example illustrates how enthalpy, entropy, and temperature collectively influence whether a chemical reaction proceeds on its own.
Other exercises in this chapter
Problem 49
Describe how a system's entropy changes if the system becomes more disordered during a process.
View solution Problem 50
Decide Does the entropy of a system increase or decrease when you disolve a cube of sugar in a cup of tea? Define the system, and explain your answer.
View solution Problem 54
How does the chemical potential energy of a system change during an endothermic reaction?
View solution Problem 55
How does the chemical potential energy of a system change during an endothermic reaction?
View solution