### Understanding Chemical EquilibriumAt a certain temperature, the equilibrium constant \( K \) for the following reaction is 0.0041:\[ \text{NO}_3(g) + \text{NO}(g) \rightleftharpoons 2\text{NO}_2(g) \]Use this information to complete the following table.#### Question 1**Suppose a 18. L reaction vessel is filled with 0.61 mol of NO\(_2\). What can you say about the composition of the mixture in the vessel at equilibrium?**- There will be very little NO\(_3\) and NO. (Selected)- There will be very little NO\(_2\).- Neither of the above is true.#### Question 2**What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits.**\[ 2\text{NO}_2(g) \rightleftharpoons \text{NO}_3(g) + \text{NO}(g) \]\( K = \) [ ]#### Question 3**What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits.**\[ 3\text{NO}_3(g) + 3\text{NO}(g) \rightleftharpoons 6\text{NO}_2(g) \]\( K = \) [ ]### Explanation of the Graphs and Equilibrium Table- The first section presents a scenario where the reaction occurs in an 18-liter vessel with an initial amount of 0.61 mol of NO\(_2\). The equilibrium state is described, and the student is asked about the composition of this mixture.- The second part of the table requires the calculation of equilibrium constants for two given reactions derived from the original reaction, emphasizing the importance of rounding to two significant digits.### Key Points- **Equilibrium Constant (K):** A measure of the ratio of the concentrations of the products to the reactants at equilibrium.- **Significant Digits:** Ensure calculations are accurate and rounded appropriately.- **Reaction Direction:** Understanding how the equilibrium position changes with different reaction directions.

At a certain temperature, the equilibrium constant K for the following reaction is 0.0041: NO3(g) + NO(g) 2 NO₂(g) Use this information to complete the following table. There will be very little NO3 and NO. Suppose a 18. L reaction vessel is filled with 0.61 mol of NO₂. What can you say about the composition of the mixture in the vessel at equilibrium? There will be very little NO₂. Neither of the above is true. What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits. K = 0 2 NO₂(9) NO3(9)+NO (9) What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits. K = 3 NO3(9)+3NO(g) 6 NO₂(9) X 3 ?

At a certain temperature, the equilibrium constant K for the following reaction is 0.0041: NO3(g) + NO(g) 2 NO₂(g) Use this information to complete the following table. There will be very little NO3 and NO. Suppose a 18. L reaction vessel is filled with 0.61 mol of NO₂. What can you say about the composition of the mixture in the vessel at equilibrium? There will be very little NO₂. Neither of the above is true. What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits. K = 0 2 NO₂(9) NO3(9)+NO (9) What is the equilibrium constant for the following reaction? Round your answer to 2 significant digits. K = 3 NO3(9)+3NO(g) 6 NO₂(9) X 3 ?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter13: Chemical Equilibrium

Section: Chapter Questions

Problem 117CP: A mixture of N2, H2, and NH3 is at equilibrium [according to the equationN2(g)+3H2(g)2NH3(g)] as...

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During an experiment with the Haber process, a researcher put 1 mol N2 and 1 mol H2 into a reaction vessel to observe the equilibrium formation of ammonia, NH3. N2(g)+3H2(g)2NH3(g) When these reactants come to equilibrium, assume that x mol H2 react. How many moles of ammonia form?
In Section 13.1 of your text, it is mentioned that equilibrium is reached in a closed system. What is meant by the term closed system. and why is it necessary to have a closed system in order for a system to reach equilibrium? Explain why equilibrium is not reached in an open system.
A gaseous material XY(g) dissociates to some extent to produce X(g) and Y(g): XY(g)X(g)+Y(g) A 2.00-g sample of XY (molar mass = 165 g/mol) is placed in a container with a movable piston at 25C. The pressure is held constant at 0.967 atm. As XY begins to dissociate, the piston moves until 35.0 mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of K for this reaction of 25C.
Consider the following equilibrium: COBr2(g) CO(g) + Br2(g)Kc = 0.190 at 73 C (a) A 0.50 mol sample of COBr2 is transferred to a 9.50-L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species. (b) The volume of the container is decreased to 4.5 L and the system allowed to return to equilibrium. Calculate the new equilibrium concentrations. (Hint: The calculation will be easier if you view this as a new problem with 0.5 mol of COBr2 transferred to a 4.5-L flask.) (c) What is the effect of decreasing the container volume from 9.50 L to 4.50 L?
For the equilibrium 2 SO2(g) + O2(g) 2 SO3(g) Kc = 245 (at 1000 K) the equilibrium concentrations are [SO2] = 0.102, [O2] = 0.0132, and [SO3] = 0.184. The concentration of SO2 is suddenly doubled. Show that the forward reaction takes place to reach a new equilibrium.
The following equilibrium is established in a closed container: C(s)+O2(g)CO2(g)H=393kJmol1 How does the equilibrium shift in response to each of the following stresses? (a) The quantity of solid carbon is increased. (b) A small quantity of water is added, and CO2 dissolves in it. (c) The system is cooled. (d) The volume of the container is increased.
At room temperature, the equilibrium constant Kc for the reaction 2 NO(g) ⇌ N2(g) + O2(g) is 1.4 × 1030. Is this reaction product-favored or reactant-favored? Explain your answer. In the atmosphere at room temperature the concentration of N2 is 0.33 mol/L, and the concentration of O2 is about 25% of that value. Calculate the equilibrium concentration of NO in the atmosphere produced by the reaction of N2 and O2. How does this affect your answer to Question 11?
At 2300 K the equilibrium constant for the formation of NO(g) is 1.7 103. N2(g) + O2(g) 2 NO(g) (a) Analysis shows that the concentrations of N2 and O2 are both 0.25 M, and that of NO is 0.0042 M under certain conditions. Is the system at equilibrium? (b) If the system is not at equilibrium, in which direction does the reaction proceed? (c) When the system is at equilibrium, what are the equilibrium concentrations?
1’he reaction in Exercise 12.33 was repeated. This time, the reaction began when only NO was injected into the reaction container. 110.200 mol L_l NO was injected, what were the equilibrium concentrations of all species? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) ^2 NO K = 4.1 X 10~4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L"' ofO2, what were the equilibrium concentrations of all species?
. Consider an equilibrium mixture consisting of H2O(g), CO(g). H2(g), and CO2(g) reacting in a closed vessel according to the equation H2O(g)+CO(g)H2(g)+CO2(g)a. You add more H2O to the flask. How does the new equilibrium concentration of each chemical compare to its origin al equilibrium concentration after equilibrium is re-established? Justify your answer. b. You add more H2to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is re-established? Justify your answer.
Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation: A(g)+B(g)C(g)+D(g) a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more D to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.
Consider the following reactions at some temperature: 2NOCl(g)2NO(g)+Cl2(g)K=1.6105 2NO(g)N2(g)+O2(g)K=11031 For each reaction, assume some quantities of the reactants were placed in separate containers and allowed to come to equilibrium. Describe the relative amounts of reactants and products that would be present at equilibrium. At equilibrium, which is faster, the forward or reverse reaction in each case?