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Chapter 12: Problem 101

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.

Short Answer

Expert verified
The equilibrium concentrations of NOCl, NO, and Cl鈧� are 0.074 mol/L, 1.926 mol/L, and 0.963 mol/L, respectively.

Step by step solution

01

Write the balanced reaction equation

The balanced reaction equation is given as: \[2\mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)\]
02

Set up an ICE table

Using the given information, we can set up an ICE table to keep track of the concentrations: | | NOCl | NO | Cl鈧� | |---------|-------|------|-------| | 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.
03

Write the equilibrium expression

The equilibrium constant K expression is: \[K = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
04

Substitute equilibrium concentrations into the expression

Substitute equilibrium concentrations from the ICE table into the equilibrium constant expression and the given K value: \[1.6 \times 10^{-5} = \frac{(2- 2x)^2(1-x)}{(2x)^2}\]
05

Solve for x

Solve for x: \[x = 0.037\]
06

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 Cl鈧� = 1 - x = 1 - 0.037 = 0.963 mol/L So, the equilibrium concentrations of NOCl, NO, and Cl鈧� are 0.074 mol/L, 1.926 mol/L, and 0.963 mol/L, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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,
  • the equilibrium constant expression is \[K = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
The K value indicates the extent of the reaction; a very small K, like in this problem with \(K = 1.6 \times 10^{-5}\), suggests that the reactants are predominant at equilibrium.
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.
  • **Initial**: Enter the initial concentrations of reactants and products. If not provided, assume it鈥檚 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鈥檛 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.
Understanding this concept helps us grasp why reactions don't always proceed to completion and how conditions can affect the position of equilibrium.
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:
  • \[2\mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)\]
This equation tells us:
  • 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.
Understanding and correctly balancing the reaction equation is crucial for successful equilibrium calculations and predictions.

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Most popular questions from this chapter

At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}:\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 mol/L. Calculate the value of \(K\) for this reaction.

Nitrogen gas \(\left(\mathrm{N}_{2}\right)\) reacts with hydrogen gas \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) At \(200^{\circ} \mathrm{C}\) in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 atm of hydrogen gas. At equilibrium,the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the \(K_{\mathrm{p}}\) value for this reaction.

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 \(250 .^{\circ} \mathrm{C}\) to vaporize the methanol. Over time, the methanol vapor decomposed by the \(10=\) following reaction:$$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$.After the system has reached equilibrium, a tiny hole is drilled in the side of the flask allowing gaseous compounds to effuse out of the flask. Measurements of the effusing gas show that it contains 33.0 times as much \(\mathrm{H}_{2}(g)\) as \(\mathrm{CH}_{3} \mathrm{OH}(g) .\) Calculate \(K\) for this reaction at \(250 .^{\circ} \mathrm{C}\).

For the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g), K_{\mathrm{p}}=0.25\) at a certain temperature. If 0.040 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is reacted initially, calculate the equilibrium partial pressures of \(\mathrm{NO}_{2}(g)\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\).

At \(25^{\circ} \mathrm{C},\) gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that \(12.5 \%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is 0.900 atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.
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