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Chapter 13: Problem 103

At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas are placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to \(\mathrm{NO}\) and \(\mathrm{O}_{2} :\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 \(\mathrm{mol} / \mathrm{L}\) . Calculate the value of \(K\) for this reaction.

Short Answer

Expert verified
The equilibrium constant, \(K\), for this reaction at the given temperature is approximately 0.939.

Step by step solution

01

Calculating initial concentrations

First, let's calculate the initial concentration of \(\mathrm{NO}_{2}\) by dividing the given moles (8.1 mol) by the volume of the container (3.0 L): \[\mathrm{NO}_{2} \text{ initial concentration} = \frac{8.1 \text{ moles}}{3.0 \text{ L}} = 2.7 \frac{\text{mol}}{\text{L}}\]
02

Setting up changes in concentration

Let's use the stoichiometric coefficients in the balanced chemical equation to understand the changes in concentrations during the reaction. At equilibrium, let the decrease in concentration of \(\mathrm{NO}_{2}\) be \(x\). Since two moles of \(\mathrm{NO}_{2}\) are required to produce two moles of \(\mathrm{NO}\) and one mole of \(\mathrm{O}_{2}\), the concentration of \(\mathrm{NO}_{2}\) will decrease by \(x\) molar, while the concentration of \(\mathrm{NO}\) will increase by \(x\) molar, and the concentration of \(\mathrm{O}_{2}\) will increase by \(\frac{x}{2}\) molar.
03

Calculating x

Given the equilibrium concentration of \(\mathrm{NO}(g)\) as 1.4 M, we can find the value of \(x\): \(x = \text{equilibrium concentration of } \mathrm{NO} = 1.4 \frac{\text{mol}}{\text{L}}\)
04

Finding equilibrium concentrations

Now that we have found the value of \(x\), we can find the equilibrium concentrations for all reactants and products: \[\text{Equilibrium concentration of } \mathrm{NO}_{2} = 2.7 \frac{\text{mol}}{\text{L}} - x = 2.7 - 1.4 = 1.3 \frac{\text{mol}}{\text{L}}\] \[\text{Equilibrium concentration of } \mathrm{NO} = x = 1.4 \frac{\text{mol}}{\text{L}}\] \[\text{Equilibrium concentration of } \mathrm{O}_{2} = \frac{x}{2} = \frac{1.4}{2} = 0.7 \frac{\text{mol}}{\text{L}}\]
05

Calculating the equilibrium constant, K

Using the equilibrium concentrations, we can write the expression for the equilibrium constant, \(K\): \[K = \frac{[\mathrm{NO}]^{2}[\mathrm{O}_{2}]}{[\mathrm{NO}_{2}]^{2}}\] Now, substitute the equilibrium concentrations into the expression: \[K = \frac{(1.4)^{2}(0.7)}{(1.3)^{2}} = \frac{1.586}{1.69}\] After performing the calculation: \[K \approx 0.939\]
06

Final Answer

The equilibrium constant, \(K\), for this reaction at the given temperature is approximately 0.939.

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Most popular questions from this chapter

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. $P_{\mathrm{NO}}=0.0062 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.51 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.18 \mathrm{atm}$

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: $[\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\( \)\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} M,$ and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]-2.9 \times 10^{-3} \mathrm{M} .\) Calculate the value of \(K\) for the reaction at this temperature.

At a particular temperature, \(K_{\mathrm{p}}=1.00 \times 10^{2}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)$$ If 2.00 atm of \(\mathrm{H}_{2}(g)\) and 2.00 atm of \(\mathrm{I}_{2}(g)\) are introduced into a \(1.00-\mathrm{L}\) container, calculate the equilibrium partial pressures of all species.

At a particular temperature, 8.0 moles of \(\mathrm{NO}_{2}\) is placed into a 1.0 -L container and the \(\mathrm{NO}_{2}\) dissociates by the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) is 2.0 \(\mathrm{M}\) . Calculate \(K\) for this reaction.

Consider the reaction $$\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q)$$ How will the equilibrium position shift if a. water is added, doubling the volume? b. \(\operatorname{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? [Fe(OH) \(_{3}\) is insoluble. \(]\) d. Fe(NO \(_{3} )_{3}(a q)\) is added?
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