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Chapter 13: Problem 45

A 1.00-L flask was filled with \(2.00\) mol gaseous \(\mathrm{SO}_{2}\) and \(2.00 \mathrm{~mol}\) gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that \(1.30\) mol 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 for the reaction \( \mathrm{SO}_{2}(g) + \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) + \mathrm{NO}(g) \) under the given conditions is approximately 3.42.

Step by step solution

01

Set up the ICE table

An ICE table is used to organize the initial concentrations, changes, and equilibrium concentrations of each species in the reaction. The table looks like this: | | SO2 | NO2 | SO3 | NO | |:---------:|:---:|:---:|:---:|:---:| | Initial | 2 | 2 | 0 | 0 | | Change | -x | -x | +x | +x | | Equilibrium|2-x | 2-x | x | x | Notice how the changes in the concentration of SO2 and NO2 are negative (since they are being used up in the reaction) and the changes in the concentration of SO3 and NO are positive (since they are being produced).
02

Calculate the equilibrium concentrations

We are given the amount of NO at equilibrium, which is 1.30 mol. Since the volume of the flask is 1.00 L, we can find the concentration of NO at equilibrium: \[ \mathrm{[NO]}_{eq}= \frac{1.3 \mol}{1.0 \mathrm{L}} = 1.3 \mathrm{M} \] Since the change in concentration for NO is +x, and the equilibrium concentration is 1.3 M, we can find the value of x: \( x = 1.3 \mathrm{M} \) Now we can find the equilibrium concentrations of the other species by plugging in the value of x into the equilibrium row of our ICE table: \[ [\mathrm{SO}_2]_{eq} = 2 - x = 2 - 1.3 = 0.7 \mathrm{M} \] \[ [\mathrm{NO}_2]_{eq} = 2 - x = 2 - 1.3 = 0.7 \mathrm{M} \] \[ [\mathrm{SO}_3]_{eq} = x = 1.3 \mathrm{M} \]
03

Calculate the equilibrium constant, K

Now that we have the equilibrium concentrations of all species, we can calculate the equilibrium constant, K, using this equation: \[ K = \frac{[\mathrm{SO}_3][\mathrm{NO}]}{[\mathrm{SO}_2][\mathrm{NO}_2]} \] Plug in the equilibrium concentrations: \[ K = \frac{(1.3)(1.3)}{(0.7)(0.7)} \] Now, compute the equilibrium constant, K: \[ K = \approx 3.42 \] So, the equilibrium constant for this reaction is approximately 3.42.

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Most popular questions from this chapter

Consider the following reactions. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) and \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\) List two property differences between these two reactions that relate to equilibrium.

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\), equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

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 \(2.0 \mathrm{~mol} \mathrm{NO}\) and \(1.0 \mathrm{~mol} \mathrm{Cl}_{2}\) are placed into a \(1.0\) - \(\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Consider the following reaction at a certain temperature: $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) $$ An equilibrium mixture contains \(1.0 \mathrm{~mol} \mathrm{Fe}, 1.0 \times 10^{-3} \mathrm{~mol} \mathrm{O}_{2}\), and \(2.0 \mathrm{~mol} \mathrm{Fe}_{2} \mathrm{O}_{3}\) all in a 2.0-L container. Calculate the value of \(K\) for this reaction.
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