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Chapter 13: Problem 32

At a particular temperature a \(2.00-\mathrm{L}\) flask at equilibrium contains \(2.80 \times 10^{-4}\) mole of \(\mathrm{N}_{2}, 2.50 \times 10^{-5}\) mole of \(\mathrm{O}_{2},\) and \(2.00 \times 10^{-2}\) mole of $\mathrm{N}_{2} \mathrm{O}\( . Calculate \)K$ at this temperature for the reaction $$2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{N}_{2} \mathrm{O}(g)$$ If $\left[\mathrm{N}_{2}\right]=2.00 \times 10^{-4} M,\left[\mathrm{N}_{2} \mathrm{O}\right]=0.200 M,\( and \)\left[\mathrm{O}_{2}\right]=\( \)0.00245 M,$ does this represent a system at equilibrium?

Short Answer

Expert verified
The equilibrium constant, \(K\), for the given reaction is approximately \(512138.4\). With the new concentrations (\([N_{2}] = 2.00 \times 10^{-4}M\), \([N_{2}O] = 0.200M\), and \([O_{2}] = 0.00245M)\), the reaction quotient, \(Q\), is approximately \(80387771.5\). Since \(Q > K\), the reaction will shift to the left (towards reactants) to reach equilibrium, meaning the given concentrations represent a system that is not at equilibrium.

Step by step solution

01

Write the expression for K

We're given the balanced chemical equation: \(2N_{2}(g) + O_{2}(g) \rightleftharpoons 2N_{2}O(g)\). According to the law of mass action, the equilibrium constant K is given by: \(K = \frac{[N_{2}O]^2}{[N_{2}]^2 * [O_{2}]}\)
02

Substitute equilibrium concentrations and calculate K

Given the equilibrium concentrations for each species as follows: - \([N_{2}] = 2.80 \times 10^{-4}\, mole/L\) - \([O_{2}] = 2.50 \times 10^{-5}\, mole/L\) - \([N_{2}O] = 2.00 \times 10^{-2}\, mole/L\) Now, we can substitute these values into the K expression: \(K = \frac{(2.00 \times 10^{-2})^2}{(2.80 \times 10^{-4})^2 * (2.50 \times 10^{-5})}\) Calculating K, we get: \(K \approx 512138.4\)
03

Substitute the new concentrations in Q (Reaction Quotient) expression

The new concentrations given are: - \([N_{2}] = 2.00 \times 10^{-4}\, M\) - \([O_{2}] = 0.00245 \mathrm{M}\)` - \([N_{2}O] = 0.200 \mathrm{M}\) The expression for Q is the same as the expression for K: \(Q = \frac{[N_{2}O]^2}{[N_{2}]^2 * [O_{2}]} = \frac{(0.200)^2}{(2.00 \times 10^{-4})^2 * (0.00245)}\) Calculating Q, we get: \(Q \approx 80387771.5\)
04

Compare Q and K to determine if the system is at equilibrium

Now we need to compare the calculated values for Q and K: - If \(Q > K\), the reaction will shift to the left (towards reactants) to reach equilibrium. - If \(Q < K\), the reaction will shift to the right (towards products) to reach equilibrium. - If \(Q = K\), the system is at equilibrium. In this case, we have: \(Q \approx 80387771.5\) \(K \approx 512138.4\) Since \(Q > K\), the reaction will shift to the left (towards reactants) to reach equilibrium. Thus, the given concentrations (\([N_{2}]=2.00 \times 10^{-4}M\), \([N_{2}O]=0.200M\), and \([O_{2}]=0.00245M\)) represent a system that is not at equilibrium.

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Most popular questions from this chapter

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of 4.5 \(\mathrm{atm}\) is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of 9.0 \(\mathrm{atm}\) is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?

The equilibrium constant \(K_{\mathrm{p}}\) is \(2.4 \times 10^{3}\) at a certain temperature for the reaction $$2 \mathrm{NO}(g) \leftrightharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those not at equilibrium, in which direction will the system shift? a. $P_{\mathrm{NO}}=0.012 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.11 \mathrm{atm}, P_{\mathrm{O}_{2}}=2.0 \mathrm{atm}$ b. $P_{\mathrm{NO}}=0.0078 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.36 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.67 \mathrm{atm}$ c. $P_{\mathrm{NO}}=0.0062 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.51 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.18 \mathrm{atm}$

A 1.00-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.

For the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ at \(600 . \mathrm{K}\) , the equilibrium constant, \(K_{\mathrm{p}},\) is \(11.5 .\) Suppose that 2.450 \(\mathrm{g} \mathrm{PCl}_{5}\) is placed in an evacuated 500 -mL bulb, which is then heated to \(600 . \mathrm{K}\) . a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of PCl_ at equilibrium?

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g) :\) $$\mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g)$$ A 2.00 -g sample of \(\mathrm{XY}\) (molar mass $=165 \mathrm{g} / \mathrm{mol} )$ is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\) . The pressure is held constant at 0.967 \(\mathrm{atm} .\) As \(\mathrm{XY}\) begins to dissociate, the piston moves until 35.0 mole percent of the original \(\mathrm{XY}\) 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 $25^{\circ} \mathrm{C}$ .
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