Answer :
Sure! Let's tackle each question step-by-step:
### 1. Discussion on Properties ΔH and ΔS and Temperature Effects
a. Which property, ΔH or ΔS, favors spontaneity in the forward direction?
- ΔH (Enthalpy): A negative value of ΔH would favor spontaneity in the forward direction because it indicates that the reaction is exothermic (releasing energy), which generally favors spontaneity.
- ΔS (Entropy): A positive value of ΔS helps favor spontaneity in the forward direction because it indicates an increase in disorder, which nature tends to favor.
Given that for this reaction:
- ΔH = +137 kJ/mol
- ΔS = +12.4 J/K
Here, ΔH is positive, which does not favor spontaneity because it requires energy input (endothermic). However, ΔS is positive, which does favor spontaneity in the forward direction.
b. Which favors spontaneity in the reverse direction?
For the reverse direction to be spontaneous:
- ΔH should be negative (indicating an exothermic reaction in the reverse direction).
- ΔS would need to be negative, as a decrease in entropy would favor the reverse reaction.
However, in this reaction, since ΔH is positive and ΔS is positive, it inherently disfavors reverse spontaneity due to the positive ΔH unless temperature conditions are considered.
c. Does the reaction become more spontaneous at high temperature or low temperature?
The spontaneity of the reaction can be determined by the Gibbs Free Energy equation: \( \Delta G = \Delta H - T \Delta S \).
Given that for this reaction:
- ΔH = +137 kJ/mol = 137,000 J/mol
- ΔS = +12.4 J/K
At higher temperatures, \( T \Delta S \) will become larger.
Since ΔS is positive, increasing temperature (T) will make \( T \Delta S \) larger in the \( \Delta G = \Delta H - T \Delta S \) equation. If \( T \Delta S \) becomes large enough, it can outweigh ΔH, making ΔG negative and thereby making the reaction more spontaneous at high temperatures.
In summary:
- Positive ΔS and positive ΔH indicate that the reaction could become more spontaneous at higher temperatures in the forward direction.
### 2. Calculate ΔG at 20.0°C and Determine Spontaneity
Given:
- ΔH = +137 kJ/mol = 137,000 J/mol
- ΔS = +12.4 J/K
- Temperature = 20.0 °C = 20.0 + 273.15 K = 293.15 K
Calculate ΔG using the formula:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Substituting the values:
[tex]\[ \Delta G = 137000 \, \text{J/mol} - (293.15 \, \text{K} \times 12.4 \, \text{J/K}) \][/tex]
[tex]\[ \Delta G = 137000 - 3635.06 \][/tex]
[tex]\[ \Delta G = 133364.94 \, \text{J/mol} \][/tex]
Since ΔG is positive, the reaction is not spontaneous in the forward direction at 20.0°C.
- If ΔG < 0: The reaction is spontaneous in the forward direction.
- If ΔG > 0: The reaction is spontaneous in the reverse direction.
In this case:
- Since \( \Delta G = 133364.94 \, \text{J/mol} \) (which is greater than 0), the reaction is spontaneous in the reverse direction at 20.0°C.
Thus, for the reaction with ΔH = +137 kJ/mol and ΔS = +12.4 J/K at 20.0°C, the reaction is not spontaneous in the forward direction and is spontaneous in the reverse direction.
### 1. Discussion on Properties ΔH and ΔS and Temperature Effects
a. Which property, ΔH or ΔS, favors spontaneity in the forward direction?
- ΔH (Enthalpy): A negative value of ΔH would favor spontaneity in the forward direction because it indicates that the reaction is exothermic (releasing energy), which generally favors spontaneity.
- ΔS (Entropy): A positive value of ΔS helps favor spontaneity in the forward direction because it indicates an increase in disorder, which nature tends to favor.
Given that for this reaction:
- ΔH = +137 kJ/mol
- ΔS = +12.4 J/K
Here, ΔH is positive, which does not favor spontaneity because it requires energy input (endothermic). However, ΔS is positive, which does favor spontaneity in the forward direction.
b. Which favors spontaneity in the reverse direction?
For the reverse direction to be spontaneous:
- ΔH should be negative (indicating an exothermic reaction in the reverse direction).
- ΔS would need to be negative, as a decrease in entropy would favor the reverse reaction.
However, in this reaction, since ΔH is positive and ΔS is positive, it inherently disfavors reverse spontaneity due to the positive ΔH unless temperature conditions are considered.
c. Does the reaction become more spontaneous at high temperature or low temperature?
The spontaneity of the reaction can be determined by the Gibbs Free Energy equation: \( \Delta G = \Delta H - T \Delta S \).
Given that for this reaction:
- ΔH = +137 kJ/mol = 137,000 J/mol
- ΔS = +12.4 J/K
At higher temperatures, \( T \Delta S \) will become larger.
Since ΔS is positive, increasing temperature (T) will make \( T \Delta S \) larger in the \( \Delta G = \Delta H - T \Delta S \) equation. If \( T \Delta S \) becomes large enough, it can outweigh ΔH, making ΔG negative and thereby making the reaction more spontaneous at high temperatures.
In summary:
- Positive ΔS and positive ΔH indicate that the reaction could become more spontaneous at higher temperatures in the forward direction.
### 2. Calculate ΔG at 20.0°C and Determine Spontaneity
Given:
- ΔH = +137 kJ/mol = 137,000 J/mol
- ΔS = +12.4 J/K
- Temperature = 20.0 °C = 20.0 + 273.15 K = 293.15 K
Calculate ΔG using the formula:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Substituting the values:
[tex]\[ \Delta G = 137000 \, \text{J/mol} - (293.15 \, \text{K} \times 12.4 \, \text{J/K}) \][/tex]
[tex]\[ \Delta G = 137000 - 3635.06 \][/tex]
[tex]\[ \Delta G = 133364.94 \, \text{J/mol} \][/tex]
Since ΔG is positive, the reaction is not spontaneous in the forward direction at 20.0°C.
- If ΔG < 0: The reaction is spontaneous in the forward direction.
- If ΔG > 0: The reaction is spontaneous in the reverse direction.
In this case:
- Since \( \Delta G = 133364.94 \, \text{J/mol} \) (which is greater than 0), the reaction is spontaneous in the reverse direction at 20.0°C.
Thus, for the reaction with ΔH = +137 kJ/mol and ΔS = +12.4 J/K at 20.0°C, the reaction is not spontaneous in the forward direction and is spontaneous in the reverse direction.
