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Chapter 12: Problem 59

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$. If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 - \(\mathrm{L}\) vessel, calculate the equilibrium concentrations of all species.

Short Answer

Expert verified
At the given temperature, the equilibrium concentrations of CO2, CO, and O2 are approximately \( CO2 = 0.40 M - 2x, CO = 2x, O2 = x\). Solving the equilibrium constant equation, \( 2.0 * 10^{-6} = \frac{(2x)^2 * x}{(0.40 - 2x)^2}\), we find \(x \approx 3.97 * 10^{-5}\). Therefore, the equilibrium concentrations of CO2, CO, and O2 are approximately 0.3992 M, 7.94 * 10^{-5} M, and 3.97 * 10^{-5} M respectively.

Step by step solution

01

Find the initial concentrations of the species

In this step, we calculate the initial concentration of CO2. We are given the amount of CO2 in moles (2.0 moles) and the volume of the vessel (5.0 L). To find the initial concentration, we use the formula: Concentration = (moles of substance) / (volume of solution in liters) We find the initial concentration of CO2: CO2_initial = (2.0 moles) / (5.0 L) = 0.40 M Since no CO and O2 are mentioned, their initial concentrations will be 0.
02

Set up the ICE table

An initial-change-equilibrium (ICE) table will help us analyze the changes in concentrations of the reactants and products as the reaction proceeds. We will use x as the change in concentration of CO, CO2, and O2 from their initial values. | | CO2 | CO | O2 | |---------|--------------|------------|--------- | | Initial | 0.40 M | 0 M | 0 M | | Change | -2x | +2x | +x | | Equilibrium | 0.40 M - 2x | 0 M + 2x | 0 M + x |
03

Write the equilibrium constant expression

Write the equilibrium constant expression for the given reaction using the concentrations at equilibrium from the ICE table. The equation for the equilibrium constant (K) is: K = [CO]^2 * [O2] / [CO2]^2 Substitute the equilibrium concentrations from the ICE table: K = [(2x)]^2 * [(x)] / [(0.40 - 2x)]^2 Given K = 2.0 * 10^(-6), substitute K in the equation: 2.0 * 10^(-6) = (4x^2 * x) / (0.40 - 2x)^2
04

Solve for x

To solve for x, the change in concentration from the initial values, we can first simplify the equation: 2.0 * 10^(-6) * (0.40 - 2x)^2 = 4x^3 Now we need to solve this equation for x. The equation is a bit complicated to solve algebraically, so we will use an iterative or trial and error method to find the approximate value of x.
05

Find the equilibrium concentrations

After obtaining a value for x, we can find the equilibrium concentrations of CO2, CO, and O2 by substituting x back into the ICE table. CO2_equilibrium = 0.40 M - 2x CO_equilibrium = 0 M + 2x O2_equilibrium = 0 M + x You can compute the final equilibrium concentrations by substituting the numerical value you found for x in the previous step.

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Most popular questions from this chapter

At a particular temperature, \(K=4.0 \times 10^{-7}\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$,In an experiment, 1.0 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is placed in a 10.0 -L vessel. Calculate the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) when this reaction reaches equilibrium.

Hydrogen for use in ammonia production is produced by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\text { Ni catalyst }}{=\frac{750^{\circ} \mathrm{C}}{7}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

Peptide decomposition is one of the key processes of digestion, where a peptide bond is broken into an acid group and an amine group. We can describe this reaction as follows: \(\text { Peptide }(a q)+\mathrm{H}_{2} \mathrm{O}(t) \rightleftharpoons \text { acid group }(a q)+\text { amine group }(a q)\)

Lexan is a plastic used to make compact discs, eyeglass lenses, and bulletproof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right),\) an extremely poisonous gas. Phosgene decomposes by the reaction,$$\operatorname{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$,for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.

A \(1.00-\mathrm{L}\) flask was filled with 2.00 moles of gaseous \(\mathrm{SO}_{2}\) and 2.00 moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$, occurs under these conditions. Calculate the value of the equilibrium constant, \(K\), for this reaction.
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