The F-F bond in F 2 is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; K p = 7.83 at 1500 K for the reaction F 2 (g) ⇌ 2 F(g). (a) If the equilibrium partial pressure of F 2 molecules at 1500 K is 0.200 atm, what is the equilibrium partial pressure of F atoms in atm?

Hello everyone today. We are being given the following problem and asked to solve for it. So it says the equilibrium constant K. P. For the oxidation of nitric oxide is 1152 at 1 50 degrees Celsius. Were then given the following chemical reaction In that reaction mixture at equilibrium oxygen has a partial pressure of .345 atmospheres. While the partial pressure of nitrogen dioxide is .225 atmospheres were then asked to calculate the partial pressure of nitric oxide at equilibrium. So the first thing we want to do does he want to calculate, we want to write out our K. P. For the expression below. So R. K. P. It's going to be equal to the pressures of our products over reactivates. And so putting this together, We have our pressure of our only product which is nitrogen dioxide in 02. And since there is a two coefficient we're going to make this exponent into a. two and our denominator we have to react. It's present. We have nitric oxide and so we have the pressure of N. O. Nitric oxide and that has a two coefficient as well. So we're going to give that an exponent of two. And lastly we're going to multiply the second reactant from by the first one which is just molecular oxygen. And so we have our K. P. Expression below. We also were given a partial pressure of oxygen and the partial pressure of pressure of nitrogen dioxide. And so ultimately we want to find the partial pressure of nitric oxide. So we're gonna label that X. And so Plugging our values in to this reaction that we just found we have our K. P. or equilibrium constant being 1,152. And that's going to equal Our partial pressure of our nitrogen dioxide. Which in the question Stem said that it was 0.225 atmospheres. And we're going to square that and then we're going to divide that by our partial pressure of nitric oxide, which we don't know. So we're just gonna say X. And we're gonna square that as well. And then our partial pressure for molecular oxygen or 02 is 0.345 atmospheres, solving for X. We get an X. is equal to 0.0113 atmospheres. In other words, this is our partial pressure of nitric oxide. I hope this helped. And until next time.

Ch.15 - Chemical Equilibrium

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McMurry 8th Edition

Ch.15 - Chemical Equilibrium

Problem 151a

Chapter 15, Problem 151a

The F-F bond in F₂ is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; K_p = 7.83 at 1500 K for the reaction F₂(g) ⇌ 2 F(g). (a) If the equilibrium partial pressure of F₂ molecules at 1500 K is 0.200 atm, what is the equilibrium partial pressure of F atoms in atm?

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Start by writing the expression for the equilibrium constant Kp for the reaction F2(g) ⇌ 2 F(g). The expression is Kp = (P_F)^2 / P_F2, where P_F is the partial pressure of F atoms and P_F2 is the partial pressure of F2 molecules.

Substitute the given values into the Kp expression. You know that Kp = 7.83 and P_F2 = 0.200 atm. So, the equation becomes 7.83 = (P_F)^2 / 0.200.

Rearrange the equation to solve for (P_F)^2. Multiply both sides by 0.200 to get (P_F)^2 = 7.83 * 0.200.

Take the square root of both sides to solve for P_F. This will give you P_F = sqrt(7.83 * 0.200).

Calculate the value of P_F using the expression from the previous step to find the equilibrium partial pressure of F atoms in atm.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (Kp)

The equilibrium constant, Kp, is a measure of the ratio of the partial pressures of products to reactants at equilibrium for a given reaction at a specific temperature. For the reaction F2(g) ⇌ 2 F(g), Kp = 7.83 indicates that at 1500 K, the ratio of the pressure of F atoms squared to the pressure of F2 is constant. Understanding Kp is essential for calculating the equilibrium concentrations or pressures of the species involved.

Partial Pressure

Partial pressure is the pressure exerted by a single component of a gas mixture. In the context of the reaction, the total pressure is the sum of the partial pressures of F2 and F atoms. By knowing the partial pressure of F2 at equilibrium, one can use the equilibrium constant to find the partial pressures of the other species, which is crucial for solving the problem.

Stoichiometry of the Reaction

Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. In the given reaction, the stoichiometry indicates that one mole of F2 produces two moles of F atoms. This relationship is vital for determining how changes in the concentration or pressure of one species affect the others at equilibrium, allowing for the calculation of the equilibrium partial pressure of F atoms based on the given conditions.

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At 1000 K, Kp = 2.1 * 106 and ΔH° = - 107.7 kJ for the reaction H21g2 + Br21g2 ∆ 2 HBr1g2.(b) For the equilibrium in part (a), each of the following changes will increase the equilibrium partial pressure of HBr. Choose the change that will cause the greatest increase in the pressure of HBr, and explain your choice.(ii) Adding 0.10 mol of Br2

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Textbook Question

Halogen lamps are ordinary tungsten filament lamps in which the lamp bulb contains a small amount of a halogen (often bromine). At the high temperatures of the lamp, the halogens dissociate and exist as single atoms.(c) When the WBr41g2 diffuses back toward the filament, it decomposes, depositing tungsten back onto the fila- ment. Show quantitatively that the pressure of WBr4 from part (a) will cause the reaction in part (a) to go in reverse direction at 2800 K. [The pressure of Br1g2 is still 0.010 atm.] Thus, tungsten is continually recycled from the walls of the bulb back to the filament, allow-ing the bulb to last longer and burn brighter.

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Textbook Question

The F-F bond in F₂ is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; K_p = 7.83 at 1500 K for the reaction F₂(g) ⇌ 2 F(g). (b) What fraction of the F₂ molecules dissociate at 1500 K?

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Textbook Question

The F-F bond in F₂ is relatively weak because the lone pairs of electrons on one F atom repel the lone pairs on the other F atom; K_p = 7.83 at 1500 K for the reaction F₂(g) ⇌ 2 F(g). (c) Why is the F-F bond in F₂ weaker than the Cl-Cl bond in Cl₂?

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