Problem 87
Question
At \(27^{\circ} \mathrm{C}, K_{\mathrm{P}}=1.5 \times 10^{18}\) for the reaction $$ 3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) $$ If \(0.030 \mathrm{~mol}\) of \(\mathrm{NO}\) were placed in a \(1.00 \mathrm{~L}\) vessel and this equilibrium were established, what would be the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2} \mathrm{O},\) and \(\mathrm{NO}_{2} ?\)
Step-by-Step Solution
Verified Answer
Due to the complexity of the algebra involved with high powers and large equilibrium constants, a numerical solver or iterative approach might be necessary to accurately determine the equilibrium concentrations.
1Step 1: Write the Equilibrium Expression
Write the expression for the equilibrium constant, \(K_{\mathrm{P}}\), for the given reaction: \[ K_{\mathrm{P}} = \frac{[\mathrm{N}_2\mathrm{O}][\mathrm{NO}_2]}{[\mathrm{NO}]^3} \]
2Step 2: Determine the Initial Concentrations
Determine the initial concentrations of the reactants and products before any reaction occurs. In this case, only \(\mathrm{NO}\) is present initially, with a concentration of \(0.030 \mathrm{~mol/L}\). There is no \(\mathrm{N}_2\mathrm{O}\) or \(\mathrm{NO}_2\) at the start.
3Step 3: Use ICE Table to Express Equilibrium Concentrations
Construct an ICE (Initial, Change, Equilibrium) table to indicate the initial concentrations, changes in concentrations, and final equilibrium concentrations of all species involved in the reaction. Let \(x\) represent the change in concentration of \(\mathrm{NO}\) that reacts, then the change for \(\mathrm{N}_2\mathrm{O}\) and \(\mathrm{NO}_2\) will be \(x\) each, and the change for \(\mathrm{NO}\) will be \(-3x\).
4Step 4: Substitute the Equilibrium Concentrations into the Equilibrium Expression
Substitute the expressions for the equilibrium concentrations in terms of \(x\) from the ICE table into the equilibrium constant expression and solve for \(x\).
5Step 5: Calculate Equilibrium Concentrations
Calculate the equilibrium concentrations for all species using the value of \(x\) obtained in the previous step.
6Step 6: Check the Validity of the Assumption
Verify that the assumption that allowed us to ignore \(-3x\) when compared to the initial concentration of \(\mathrm{NO}\) is valid. Specifically, the value of \(x\) should be small relative to the initial concentration to validate the approximation.
7Step 7: Present the Final Answer
If the assumption is valid, report the equilibrium concentrations of \(\mathrm{NO}\), \(\mathrm{N}_2\mathrm{O}\), and \(\mathrm{NO}_2\). If not, a more exact method, like the quadratic formula, might be needed to solve for \(x\).
Key Concepts
Equilibrium Constant (Kp)ICE Table MethodologyEquilibrium CalculationReaction Quotient
Equilibrium Constant (Kp)
Understanding the equilibrium constant, denoted as \( K_p \), is crucial for predicting the positions and extents of chemical reactions that reach a state of dynamic equilibrium. \( K_p \) reflects the ratio of the products to reactants, each raised to the power of their stoichiometric coefficients, under conditions of constant pressure.
In the given exercise, the equilibrium constant for the reaction involving nitrogen monoxide (NO), nitrogen dioxide (NO2), and dinitrogen monoxide (N2O) is \(1.5 \times 10^{18}\) at \(27^\circ\mathrm{C}\). It's a significantly large number, indicating the reaction heavily favors the formation of products under these conditions. A high \(K_p\) value implies the reaction will proceed almost to completion before reaching equilibrium, leaving very little reactant behind.
To bring it home, imagine a crowded party where nearly everyone wants to move from the noisy living room (reactant NO) to the quieter kitchen (products N2O and NO2). The equilibrium constant would be the ratio of people in the kitchen to those left in the living room, raised to their respective 'coefficients'—the number of people moving at a time.
In the given exercise, the equilibrium constant for the reaction involving nitrogen monoxide (NO), nitrogen dioxide (NO2), and dinitrogen monoxide (N2O) is \(1.5 \times 10^{18}\) at \(27^\circ\mathrm{C}\). It's a significantly large number, indicating the reaction heavily favors the formation of products under these conditions. A high \(K_p\) value implies the reaction will proceed almost to completion before reaching equilibrium, leaving very little reactant behind.
To bring it home, imagine a crowded party where nearly everyone wants to move from the noisy living room (reactant NO) to the quieter kitchen (products N2O and NO2). The equilibrium constant would be the ratio of people in the kitchen to those left in the living room, raised to their respective 'coefficients'—the number of people moving at a time.
ICE Table Methodology
The ICE table, an acronym for Initial, Change, Equilibrium, is a strategic tool for solving equilibrium problems. It systematically organizes the concentration data of reactants and products through the different stages of a reaction reaching equilibrium.
Here's a breakup of the ICE methodology:
Here's a breakup of the ICE methodology:
- Initial: Marks the concentrations before the reaction starts. Think of it as the guest list before the party begins.
- Change: Indicates how concentrations shift due to the reaction, akin to guests moving between rooms at the party.
- Equilibrium: Shows the concentrations once the reaction has settled into equilibrium—the 'new normal' guest distribution between rooms.
Equilibrium Calculation
When it comes to finding the equilibrium concentrations, a bit of algebra is needed. After using the ICE table to express the concentrations at equilibrium in terms of the change variable \(x\), we inject these expressions into the equilibrium constant equation—like pieces of a puzzle.
In our example, once \(x\) is found by rearranging and solving the equation, it tells us how much NO has reacted. A positive value of \(x\) implies a progression towards the formation of products. And just like pieces of a puzzle fitting together to showcase the bigger picture, inserting the value of \(x\) back into the equilibrium expressions from the ICE table reveals the specific concentrations of NO, N2O, and NO2 at equilibrium.
It's important to remember that solutions must be physically meaningful; negative concentrations don't exist in our chemical 'party', and not all reactions will move to the extent suggested by an initial \(x\) value. In some cases, like very small \(K_p\) values, \(x\) could be so small that it doesn't make a significant difference and can be ignored to simplify calculations.
In our example, once \(x\) is found by rearranging and solving the equation, it tells us how much NO has reacted. A positive value of \(x\) implies a progression towards the formation of products. And just like pieces of a puzzle fitting together to showcase the bigger picture, inserting the value of \(x\) back into the equilibrium expressions from the ICE table reveals the specific concentrations of NO, N2O, and NO2 at equilibrium.
It's important to remember that solutions must be physically meaningful; negative concentrations don't exist in our chemical 'party', and not all reactions will move to the extent suggested by an initial \(x\) value. In some cases, like very small \(K_p\) values, \(x\) could be so small that it doesn't make a significant difference and can be ignored to simplify calculations.
Reaction Quotient
The reaction quotient, \( Q \), is like a snapshot of the proportions of reactants and products at any point before equilibrium is reached. It has the same form as the equilibrium constant expression but uses the current concentrations instead of those at equilibrium.
In a dynamic event, like our chemical 'party', \( Q \) lets us predict which way the reaction will proceed to reach equilibrium. If \( Q < K_p \), there are too many reactants mingling about, and the reaction will proceed forward. Conversely, if \( Q > K_p \), then the party is crowded with products, and the reaction will shift in the reverse direction to make more reactants.
By calculating \( Q \), chemists can determine whether a system is already at equilibrium or which direction it must go to get there, helping drive decisions in industrial processes and chemical synthesis.
In a dynamic event, like our chemical 'party', \( Q \) lets us predict which way the reaction will proceed to reach equilibrium. If \( Q < K_p \), there are too many reactants mingling about, and the reaction will proceed forward. Conversely, if \( Q > K_p \), then the party is crowded with products, and the reaction will shift in the reverse direction to make more reactants.
By calculating \( Q \), chemists can determine whether a system is already at equilibrium or which direction it must go to get there, helping drive decisions in industrial processes and chemical synthesis.
Other exercises in this chapter
Problem 83
The reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) has \(K_{\mathrm{P}}=0.140\) at \(25^{\circ} \mathrm{C}\). In a reacti
View solution Problem 85
The following reaction in aqueous solution has \(K_{\mathrm{c}}=\) \(1 \times 10^{-85}\) at a temperature of \(25^{\circ} \mathrm{C}\). \(7 \mathrm{IO}_{3}^{-}(
View solution Problem 88
Consider the equilibrium \(2 \mathrm{NaHSO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g)\) How w
View solution Problem 93
To study the following reaction at \(20^{\circ} \mathrm{C}\), \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HNO}
View solution