Problem 101
Question
At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$.If 2.0 moles of \(\mathrm{NO}\) and 1.0 mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.
Step-by-Step Solution
Verified Answer
The equilibrium concentrations of NOCl, NO, and Cl₂ are 0.074 mol/L, 1.926 mol/L, and 0.963 mol/L, respectively.
1Step 1: Write the balanced reaction equation
The balanced reaction equation is given as: \[2\mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)\]
2Step 2: Set up an ICE table
Using the given information, we can set up an ICE table to keep track of the concentrations:
| | NOCl | NO | Cl₂ |
|---------|-------|------|-------|
| Initial | 0 | 2.0 | 1.0 |
| Change | +2x | -2x | -x |
| Equil. | 2x | 2-2x | 1-x |
In which x stands for the change in concentration.
3Step 3: Write the equilibrium expression
The equilibrium constant K expression is: \[K = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
4Step 4: Substitute equilibrium concentrations into the expression
Substitute equilibrium concentrations from the ICE table into the equilibrium constant expression and the given K value:
\[1.6 \times 10^{-5} = \frac{(2- 2x)^2(1-x)}{(2x)^2}\]
5Step 5: Solve for x
Solve for x:
\[x = 0.037\]
6Step 6: Calculate the equilibrium concentrations
Substitute the value of x back into the equilibrium row of the ICE table and calculate the equilibrium concentrations of all species:
- Equilibrium concentration of NOCl = 2x = 2(0.037) = 0.074 mol/L
- Equilibrium concentration of NO = 2 - 2x = 2 - 2(0.037) = 1.926 mol/L
- Equilibrium concentration of Cl₂ = 1 - x = 1 - 0.037 = 0.963 mol/L
So, the equilibrium concentrations of NOCl, NO, and Cl₂ are 0.074 mol/L, 1.926 mol/L, and 0.963 mol/L, respectively.
Key Concepts
Equilibrium ConstantICE TableChemical EquilibriumReaction Equation
Equilibrium Constant
The equilibrium constant, represented by the letter 'K', is a vital tool in understanding chemical equilibria.
It quantifies the ratio of the concentrations of products to reactants at equilibrium.
Each concentration is raised to the power of its respective coefficient in the balanced reaction equation.
For the given reaction,
This helps predict the direction in which a reaction needs to go to reach equilibrium.
It quantifies the ratio of the concentrations of products to reactants at equilibrium.
Each concentration is raised to the power of its respective coefficient in the balanced reaction equation.
For the given reaction,
- the equilibrium constant expression is \[K = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
This helps predict the direction in which a reaction needs to go to reach equilibrium.
ICE Table
An ICE table is a systematic way to calculate changes in concentrations of species in a chemical reaction.
"ICE" stands for Initial, Change, and Equilibrium.
In this exercise, it helps track the concentration changes from initial to equilibrium states.
"ICE" stands for Initial, Change, and Equilibrium.
In this exercise, it helps track the concentration changes from initial to equilibrium states.
- **Initial**: Enter the initial concentrations of reactants and products. If not provided, assume it’s zero for products that haven't formed yet.
- **Change**: Represent the change in concentration during the reaction by adding or subtracting variables (like \(x\)). Use stoichiometry to relate these changes based on the balanced equation.
- **Equilibrium**: Calculate the resulting concentrations by combining initial concentrations and changes. The ICE table, therefore, simplifies solving for equilibrium concentrations by visualizing the concentration shifts in each species throughout the reaction.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction.
At this point, the concentrations of all reactants and products remain constant over time.
This doesn’t mean that the concentrations are equal, but that their ratios remain constant as per the equilibrium constant. When a system reaches equilibrium:
At this point, the concentrations of all reactants and products remain constant over time.
This doesn’t mean that the concentrations are equal, but that their ratios remain constant as per the equilibrium constant. When a system reaches equilibrium:
- The macroscopic properties, such as concentration, no longer change.
- The microscopic processes continue to occur, meaning both forward and reverse reactions take place at equal rates.
Reaction Equation
The balanced reaction equation provides a complete and accurate representation of a chemical reaction.
It specifies the reactants and products involved, alongside their stoichiometry.
This is critical for writing the equilibrium expression and setting up ICE tables.
In this exercise, the balanced equation is:
It specifies the reactants and products involved, alongside their stoichiometry.
This is critical for writing the equilibrium expression and setting up ICE tables.
In this exercise, the balanced equation is:
- \[2\mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)\]
- The number of moles of each substance participating in the reaction, which guides us in calculating changes during the process using the ICE table.
- The direction of the reaction, which can be predicted by comparing initial concentrations to those at equilibrium.
Other exercises in this chapter
Problem 99
A 1.604 -g sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and 6.400 g oxygen gas are sealed into a 2.50 -L vessel at \(411^{\circ} \mathrm{C}\) and are
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A \(4.72-\mathrm{g}\) sample of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) was placed in an otherwise empty 1.00 -L flask and heated to \(250 .^{\cir
View solution Problem 102
Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressu
View solution Problem 103
At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$
View solution