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Chapter 12: Problem 112

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is 1.5 atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\) a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\) c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=\) \(1.00 \mathrm{atm}) ?\)

Short Answer

Expert verified
In short, the following results are obtained: a. \(K_{p} = 0.212\) at \(25^{\circ} C\) b. At the new equilibrium position (total pressure = 1.0 atm), the partial pressures of N2O4(g) and NO2(g) are \(P'_{N2O4} = 0.4 \ \text{atm}\) and \(P'_{NO2} = 0.6 \ \text{atm}\), respectively. c. At the new equilibrium position, 33.3% of the original N2O4(g) has dissociated.

Step by step solution

01

Calculate the initial moles of N2O4(g)

We are given that 16% (by moles) of N2O4(g) has dissociated. Since the initial moles of NO2(g) are zero, we can set up a simple proportion to calculate the initial moles of N2O4(g) before dissociation: initial moles of N2O4(g) = \(\frac{100-16}{16}\) × moles of NO2(g) We know that the total pressure of the system at equilibrium is 1.5 atm, and we can calculate the partial pressures of N2O4(g) and NO2(g) at equilibrium.
02

Calculate the partial pressures of N2O4(g) and NO2(g) at equilibrium

Let the partial pressure of NO2(g) be P_NO2 and the partial pressure of N2O4(g) be P_N2O4: P_NO2 = 2P_N2O4 (since 1 mole of N2O4 dissociates into 2 moles of NO2) The total pressure at equilibrium is given by: P_total = P_N2O4 + P_NO2 = 1.5 atm Now we have two equations with two unknowns: 1. P_NO2 = 2P_N2O4 2. P_N2O4 + P_NO2 = 1.5 atm We can solve for P_N2O4 and P_NO2.
03

Calculate the value of Kp

The equilibrium constant Kp in terms of partial pressures is given by: \[K_p = \frac{P_{NO2}^{2}}{P_{N2O4}}\] Using the values for P_N2O4 and P_NO2 calculated in Step 2, we can now calculate Kp at 25°C. #b. Calculate the equilibrium pressure of N2O4(g) and NO2(g) when the total pressure is 1.0 atm#
04

Calculate the new equilibrium pressures

When the volume of the cylinder is increased, the total pressure decreases to 1.0 atm. We can use the Kp value we calculated in Step 3 along with the new total pressure to calculate the new equilibrium pressures of N2O4(g) and NO2(g). Let the partial pressure of NO2(g) be P'_NO2 and the partial pressure of N2O4(g) be P'_N2O4: P'_NO2 = 2P'_N2O4 The new total pressure is given by: P'_total = P'_N2O4 + P'_NO2 = 1.0 atm Now we have two equations with two unknowns: 1. P'_NO2 = 2P'_N2O4 2. P'_N2O4 + P'_NO2 = 1.0 atm We can solve for P'_N2O4 and P'_NO2. #c. Calculate the percentage of dissociated N2O4(g) at the new equilibrium position#
05

Calculate the new percentage of dissociated N2O4(g)

Having found the new equilibrium pressures of N2O4(g) and NO2(g) in Step 4, we can calculate the new percentage of dissociated N2O4(g) at the equilibrium position with a total pressure of 1.00 atm: New percentage of N2O4 dissociated (by moles) = \(\frac{P'_{NO2}}{2 \times P'_{N2O4}} \times 100\) Using the values of P'_N2O4 and P'_NO2 calculated in Step 4, we can determine the percentage of dissociated N2O4(g) at the new equilibrium position.

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Most popular questions from this chapter

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$\mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(aq)+\mathrm{H}_{2}(a q)$$. =Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10},\) what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.

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.

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 \(250 .^{\circ} \mathrm{C}\).

Le Châtelier's principle is stated (Section \(12-7\) ) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease, and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.
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