Chapter 19: Problem 33
The formation of \(\mathrm{NO}(\mathrm{g})\) from its elements $$ \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g}) $$ has a standard free energy change, \(\Delta G^{\circ},\) of \(+86.58 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{p}}\) at this temperature. Comment on the connection between the sign of \(\Delta G^{\circ}\) and the magnitude of \(\bar{K}_{\mathrm{p}^{*}}\)
Short Answer
Step by step solution
Understand the Relationship
Convert Temperature to Kelvin
Rearrange the Equation
Insert Known Values
Calculate \( \ln K_p \)
Exponentiate to Find \( K_p \)
Interpret the Connection
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Free Energy Change
- If \( \Delta G^{\circ} < 0 \), the process can proceed spontaneously. It releases free energy.
- If \( \Delta G^{\circ} = 0 \), the system is at equilibrium, with no net change occurring.
- If \( \Delta G^{\circ} > 0 \), the reaction is nonspontaneous under standard conditions and requires energy input.
Equilibrium Constant
- A large \( K_p \) indicates that the products are favored at equilibrium.
- A small \( K_p \), as we find for NO formation, shows that reactants are favored, making the reaction incomplete.
- A \( K_p \) around 1 means a balanced equilibrium between reactants and products.
Gibbs Free Energy
- Processes with a decrease in Gibbs free energy \((\Delta G < 0)\) are feasible.
- Gibbs free energy balances energy changes due to enthalpy and entropy.
- In the context of chemical equilibrium, it links to the equilibrium constant \( K_p \), providing insights about reaction spontaneity and composition at equilibrium.
Reaction Spontaneity
- Spontaneous reactions have \( \Delta G^{\circ} < 0 \), meaning they happen independently.
- Nonspontaneous reactions, like the formation of NO (with \( \Delta G^{\circ} = +86.58 \text{ kJ/mol} \)), require energy input.
- The sign of \( \Delta G^{\circ} \) directly impacts the equilibrium constant, \( K_p \), and reflects on the favorability of the reaction's position.