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Chapter 13: Problem 51

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.

Short Answer

Expert verified
The equilibrium constant, \(K\), for the given reaction is 3.45 when 1.30 moles of NO are present at equilibrium.

Step by step solution

01

Write the balanced chemical equation

The balanced equation for the given reaction is: \[\mathrm{SO}_{2(g)} + \mathrm{NO}_{2(g)} \rightleftharpoons \mathrm{SO}_{3(g)} + \mathrm{NO}_{(g)}\]
02

Calculate the initial moles of each substance

According to the problem, initially there are: - 2.00 moles of \(\mathrm{SO}_{2}\) - 2.00 moles of \(\mathrm{NO}_{2}\) - 0 moles of \(\mathrm{SO}_{3}\) (Since they were not present initially) - 0 moles of \(\mathrm{NO}\) (Since they were not present initially)
03

Calculate the change in moles for each substance

At equilibrium, there are 1.30 moles of NO. So, the change in moles of each substance can be represented as follows: - \[\Delta\, \mathrm{SO}_{2} = -x\] - \[\Delta\, \mathrm{NO}_{2} = -x\] - \[\Delta\, \mathrm{SO}_{3} = x\] - \[\Delta\, \mathrm{NO} = x\] Where \(x\) is the moles of \(\mathrm{SO}_{3}\) and \(\mathrm{NO}\) formed and the moles of \(\mathrm{SO}_{2}\) and \(\mathrm{NO}_{2}\) that reacted.
04

Calculate the equilibrium moles of each substance and their concentrations

At equilibrium, the moles of each substance are: - Moles of \(\mathrm{SO}_{2} = 2.00 - x\) - Moles of \(\mathrm{NO}_{2} = 2.00 - x\) - Moles of \(\mathrm{SO}_{3} = x\) - Moles of \(\mathrm{NO} = x\) The volume of the container is 1.00 L, so the equilibrium concentrations can be found by dividing the moles by the volume of the container: - \[[\mathrm{SO}_{2}] = \frac{2.00 - x}{1.00\mathrm{L}} = 2.00 - x\] - \[[\mathrm{NO}_{2}] = \frac{2.00 - x}{1.00\mathrm{L}} = 2.00 - x\] - \[[\mathrm{SO}_{3}] = \frac{x}{1.00\mathrm{L}} = x\] - \[[\mathrm{NO}] = \frac{x}{1.00\mathrm{L}} = x\]
05

Write the expression for the equilibrium constant

The equilibrium constant, \(K,\) can be expressed as: \[K = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]}\] Substitute the equilibrium concentrations of each substance into the equation for \(K\): \[K = \frac{(x)(x)}{(2.00 - x)(2.00 - x)}\]
06

Calculate the value of \(x\)

Since we know that at equilibrium, there are 1.30 moles of NO present, we can use this information to find the value of \(x\): \[x = 1.30\]
07

Calculate the value of the equilibrium constant \(K\)

Substitute the value of \(x\) back into the equation for \(K\): \[K = \frac{(1.30)(1.30)}{(2.00 - 1.30)(2.00 - 1.30)}\] \[K = \frac{1.69}{0.49}\] Calculate the value of \(K\): \[K = 3.45\] The equilibrium constant, \(K\), for this reaction is 3.45.

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Most popular questions from this chapter

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$ b. $\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$ c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. $\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$

At a particular temperature, \(K=3.75\) for the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$ If all four gases had initial concentrations of \(0.800 M,\) calculate the equilibrium concentrations of the gases.

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCl, 0.10 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.10 mole of \)\mathrm{H}_{2} \mathrm{O}$ . b. A 2.0 -L flask contains 0.084 mole of HOCl, 0.080 mole of $\mathrm{Cl}_{2} \mathrm{O}\( , and 0.98 mole of \)\mathrm{H}_{2} \mathrm{O}$ . c. A 3.0 - flask contains 0.25 mole of HOCl, 0.0010 mole of $\mathrm{Cl}_{2} \mathrm{O},\( and 0.56 mole of \)\mathrm{H}_{2} \mathrm{O}$ .

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?

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3},\) what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?
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