Problem 44
Question
Standard enthalpy and standard entropy changes for the oxidation of ammonia at \(298 \mathrm{~K}\) are \(-382.64 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and \(-145.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. Standard Gibbs energy change for the same reaction at \(268 \mathrm{~K}\) is (a) \(-221.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-439.3 \mathrm{kJmol}^{-1}\) (d) \(-523.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-Step Solution
Verified Answer
The answer is (b) -339.3 kJ/mol.
1Step 1: Identify the formula for Gibbs energy change
The standard Gibbs energy change (ΔG°) can be calculated using the equation: \[ ΔG° = ΔH° - TΔS° \]where ΔG° is the Gibbs energy change, ΔH° is the enthalpy change, T is the temperature in Kelvin, and ΔS° is the entropy change.
2Step 2: Convert units if necessary
The values given are:
- ΔH° = -382.64 kJ/mol
- ΔS° = -145.6 J/(K·mol)
For consistent units in the formula, convert entropy from J to kJ by dividing by 1000:
- ΔS° = -0.1456 kJ/(K·mol)
3Step 3: Plug values into the Gibbs energy formula
Substitute the values into the formula:\[ ΔG° = -382.64 \, \text{kJ/mol} - 298 \, \text{K} \cdot (-0.1456 \, \text{kJ/(K·mol)}) \]
4Step 4: Calculate the change in Gibbs energy
Perform the multiplication and subtraction:\[ ΔG° = -382.64 \, \text{kJ/mol} + 43.3888 \, \text{kJ/mol} \]This simplifies to:\[ ΔG° = -339.2512 \, \text{kJ/mol} \]
5Step 5: Match the calculated Gibbs energy with options given
From the calculated value of ΔG° = -339.3 kJ/mol, match it with the given choices. The closest option is (b) -339.3 kJ/mol, which represents the Gibbs energy change at 298 K.
Key Concepts
Understanding Enthalpy ChangeDecoding Entropy ChangeOxidation of Ammonia Explained
Understanding Enthalpy Change
Enthalpy change, commonly denoted as \(ΔH\), is a crucial concept in chemistry, representing the total heat content change in a system during a reaction at constant pressure.
For any chemical reaction, it tells us whether heat is being absorbed or released.
In exothermic reactions, like the oxidation of ammonia given in the exercise, heat is released, making the \(ΔH\) value negative.
This means the products are more stable than the reactants by the release of this heat energy.
If you see a negative \(ΔH\), expect the surroundings to feel warmer, indicating an exothermic process.
Conversely, an endothermic reaction absorbs heat, indicated by a positive \(ΔH\) value.
Understanding this concept helps explain observations in reaction spontaneity as it ties closely with the Gibbs Free Energy formula.
It is one piece of the puzzle when figuring out if a reaction will occur on its own.
For any chemical reaction, it tells us whether heat is being absorbed or released.
In exothermic reactions, like the oxidation of ammonia given in the exercise, heat is released, making the \(ΔH\) value negative.
This means the products are more stable than the reactants by the release of this heat energy.
If you see a negative \(ΔH\), expect the surroundings to feel warmer, indicating an exothermic process.
Conversely, an endothermic reaction absorbs heat, indicated by a positive \(ΔH\) value.
Understanding this concept helps explain observations in reaction spontaneity as it ties closely with the Gibbs Free Energy formula.
It is one piece of the puzzle when figuring out if a reaction will occur on its own.
Decoding Entropy Change
Entropy change, represented by \(ΔS\), measures the disorder or randomness in a system.
During the oxidation of ammonia, the \(ΔS\) value given is negative, indicating that the system becomes more ordered as the reaction proceeds.
This might happen, for example, because gas molecules in the reactants form fewer or more ordered molecules in the products, decreasing randomness.
A negative entropy change isn't rare in exothermic reactions.
Or you might imagine that freeze transformations involve less disorder than gaseous states due to structured particle formation.
In spontaneous reactions, the increase or decrease in entropy is assessed along with enthalpy to calculate Gibbs Free Energy.
The relationship between this trio contributes profoundly to thermodynamics, helping predict if reactions will occur spontaneously under certain conditions.
During the oxidation of ammonia, the \(ΔS\) value given is negative, indicating that the system becomes more ordered as the reaction proceeds.
This might happen, for example, because gas molecules in the reactants form fewer or more ordered molecules in the products, decreasing randomness.
A negative entropy change isn't rare in exothermic reactions.
Or you might imagine that freeze transformations involve less disorder than gaseous states due to structured particle formation.
In spontaneous reactions, the increase or decrease in entropy is assessed along with enthalpy to calculate Gibbs Free Energy.
The relationship between this trio contributes profoundly to thermodynamics, helping predict if reactions will occur spontaneously under certain conditions.
Oxidation of Ammonia Explained
The oxidation of ammonia is an example of a combustion reaction often studied in chemistry for its energy transformations.
This process involves ammonia \((NH_3)\) reacting with oxygen \((O_2)\), resulting in products like nitrogen dioxide \((NO_2)\) and water \((H_2O)\).
The standard enthalpy and entropy changes calculated matter because they affect the energy balance and final reaction spontaneity.
A fundamental insight is that the process's exothermic nature (negative \(ΔH\)) helps gauge the release of energy and heat.
Understanding this oxidation involves deeper thermodynamic assessments where enthalpy and entropy changes interact in the Gibbs energy equation.
Ultimately, by studying such transformations, students can grasp crucial concepts in reaction dynamics and their feasibility.
These assessments decode nature's inclinations using metrics beyond superficial reaction observations.
This process involves ammonia \((NH_3)\) reacting with oxygen \((O_2)\), resulting in products like nitrogen dioxide \((NO_2)\) and water \((H_2O)\).
The standard enthalpy and entropy changes calculated matter because they affect the energy balance and final reaction spontaneity.
A fundamental insight is that the process's exothermic nature (negative \(ΔH\)) helps gauge the release of energy and heat.
Understanding this oxidation involves deeper thermodynamic assessments where enthalpy and entropy changes interact in the Gibbs energy equation.
Ultimately, by studying such transformations, students can grasp crucial concepts in reaction dynamics and their feasibility.
These assessments decode nature's inclinations using metrics beyond superficial reaction observations.
Other exercises in this chapter
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