Chapter 1: Problem 31
For a given reaction, the value of \(\Delta H\) is \(15 \mathrm{~kJ}^{-1} \mathrm{~mol}^{-1}\) and the value of \(\Delta S\) is \(51 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\). Above what temperature will this reaction be spontaneous?
Short Answer
Expert verified
The reaction is spontaneous above 294.2 K.
Step by step solution
01
Understand Spontaneity
A reaction is spontaneous when the Gibbs free energy change \( \Delta G \) is negative. The formula to calculate \( \Delta G \) is:\[ \Delta G = \Delta H - T \Delta S \] where \( \Delta H \) is enthalpy change, \( T \) is temperature in Kelvin, and \( \Delta S \) is entropy change.
02
Set Up the Inequality for Spontaneity
To find the temperature above which the reaction is spontaneous, set \( \Delta G < 0 \). This leads to the inequality:\[ 0 > \Delta H - T \Delta S \] Rearrange to find the threshold temperature \( T \):\[ T > \frac{\Delta H}{\Delta S} \]
03
Convert Units
Since \( \Delta H \) is given in \( \mathrm{kJ/mol} \) and \( \Delta S \) in \( \mathrm{J/K/mol} \), convert \( \Delta H \) to \( \mathrm{J/mol} \):\[ 15 \mathrm{~kJ/mol} = 15000 \mathrm{~J/mol} \]
04
Calculate the Threshold Temperature
Substitute the given values into the inequality:\[ T > \frac{15000 \mathrm{~J/mol}}{51 \mathrm{~J/K/mol}} \] Calculate the result:\[ T > 294.12 \mathrm{~K} \] Round up to satisfy the inequality: \[ T > 294.2 \mathrm{~K} \]
05
Interpret the Result
The reaction will be spontaneous at temperatures greater than \( 294.2 \mathrm{~K} \). Ensure to mention that this temperature is the threshold above which spontaneity is achieved.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), is a crucial aspect of understanding chemical reactions. It represents the overall heat absorbed or released during a reaction at constant pressure. Usually measured in kilojoules per mole (\( \mathrm{kJ/mol} \)), a positive \( \Delta H \) suggests the reaction absorbs heat, termed as endothermic.
Conversely, a negative \( \Delta H \) means the reaction releases heat, known as exothermic.Key points about enthalpy change:
Conversely, a negative \( \Delta H \) means the reaction releases heat, known as exothermic.Key points about enthalpy change:
- It reflects the energy change in terms of heat.
- Endothermic reactions feel cold to the touch as they absorb heat from the surroundings.
- Exothermic reactions may feel warm or hot as they release heat.
Entropy Change
Entropy change, represented as \( \Delta S \), measures the disorder or randomness in a system. The unit is usually joules per Kelvin per mole (\( \mathrm{J/K/mol} \)). A positive \( \Delta S \) indicates an increase in disorder, while a negative \( \Delta S \) suggests a decrease.Entropy is a key concept in determining the feasibility of a reaction:
- Greater disorder (higher \( \Delta S \)) usually favors spontaneity, especially at higher temperatures.
- A reaction with a positive entropy change often occurs naturally and more freely.
- Entropy change is especially significant when combined with temperature, impacting the Gibbs free energy.
Reaction Spontaneity
The concept of reaction spontaneity hinges on the Gibbs free energy change \( \Delta G \), calculated using the formula: \( \Delta G = \Delta H - T \Delta S \), where \( T \) is the temperature in Kelvin.A few critical points about spontaneity:
- The reaction is spontaneous when \( \Delta G < 0 \).
- Temperature plays a vital role in spontaneity, especially when \( \Delta H \) is positive, as seen in the given reaction.
- The inequality \( T > \frac{\Delta H}{\Delta S} \) helps determine the temperature above which the reaction becomes spontaneous.