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A reaction mixture that consisted of \(0.400 \mathrm{~mol} \mathrm{H}\) \(1.60 \mathrm{~mol} \mathrm{} \mathrm{I}_{2}\) was introduced into a \(3.00\)-L flask and heated. At equilibrium, \(60.0 \%\) of the hydrogen gas had reacted. What is the equilibrium constant \(K\) for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) at this temperature?

Short Answer

Expert verified
The equilibrium constant (K) for the reaction is 1.06.

Step by step solution

01

Calculate Moles of Reacted Hydrogen

Determine the amount of hydrogen gas reacted at equilibrium. Given that 60.0% of the initial 0.400 mol H2 has reacted, you can calculate the reacted moles as follows: (60.0/100) * 0.400 mol = 0.240 mol H2.
02

Determine Moles at Equilibrium

Since the stoichiometry of the reaction is 1:1:2 for H2:I2:HI, the moles of H2 and I2 that reacted are equal, and the moles of HI produced is twice that amount. Calculate the moles of H2, I2, and HI at equilibrium using this information:- Moles of H2 remaining = initial moles of H2 - reacted moles of H2 = 0.400 mol - 0.240 mol = 0.160 mol- Moles of I2 remaining = initial moles of I2 - reacted moles of H2 (since it's a 1:1 reaction) = 1.60 mol - 0.240 mol = 1.36 mol- Moles of HI produced = 2 * reacted moles of H2 = 2 * 0.240 mol = 0.480 mol.
03

Find Equilibrium Concentrations

The equilibrium concentrations are found by dividing the moles of each species by the volume of the flask. Since the volume of the flask is 3.00 L, we get the following concentrations:- [H2] = 0.160 mol / 3.00 L = 0.0533 M- [I2] = 1.36 mol / 3.00 L = 0.4533 M- [HI] = 0.480 mol / 3.00 L = 0.1600 M.
04

Calculate the Equilibrium Constant

The equilibrium constant, K, for the reaction can be calculated using the equilibrium concentrations in the expression for K: \[K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\].Plugging in the values, we have:\[K = \frac{(0.1600 \, \text{M})^2}{(0.0533 \, \text{M})(0.4533 \, \text{M})} = \frac{0.0256 \, \text{M}^2}{0.02416589 \, \text{M}^2} = 1.06.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Understanding chemical equilibrium is essential for interpreting how reactions proceed and the conditions under which they stabilize. It occurs when a chemical reaction and its reverse reaction proceed at the same rate, leading to a balance where the concentrations of reactants and products remain constant over time, but not necessarily equal. This state can be disrupted by changes in temperature, pressure, or the addition of other substances, illustrating the principle of Le Chatelier's, which describes how an equilibrium shifts in response to such changes.

The equilibrium constant, represented as K, is a numerical value that is unique for every reaction at a given temperature. It is derived from the balanced chemical equation of the reaction and shows the ratio of the concentration of the products to the reactants, each raised to the power of their stoichiometric coefficients. In the provided exercise, the reaction achieves equilibrium when the rates of hydrogen (H2) and iodine (I2) converting into hydrogen iodide (HI) equal the rate of HI decomposing back into H2 and I2. Calculating the equilibrium constant helps us understand the extent to which a reaction proceeds.
Reaction Stoichiometry
The concept of reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It plays a critical role in calculations such as those seen in the equilibrium constant problem. The stoichiometric coefficients from the balanced chemical equation provide the conversion factor between moles of different substances. In our case, the reaction between hydrogen and iodine to form hydrogen iodide follows the molar ratio of 1:1:2, meaning one mole of H2 reacts with one mole of I2 to produce two moles of HI.

This stoichiometric understanding is crucial for calculating the moles of reactants remaining and products formed once equilibrium is reached. For the student to effectively grasp this concept, picturing it as a simple mathematical ratio can be helpful, as well as practicing with a variety of balanced chemical equations to solidify the application of these ratios in different scenarios.
Equilibrium Concentrations
Equilibrium concentrations refer to the amounts of reactants and products present when a reaction has achieved equilibrium. These are crucial for calculating the equilibrium constant, as seen in the provided exercise. To determine these concentrations, one must take into account not only the stoichiometry of the reaction but also the volume of the reaction container. In the exercise, the remaining amount of reactants and the amount of products at equilibrium are calculated by applying the reaction stoichiometry, then converted into molar concentrations by dividing by the volume of the reaction vessel.

By understanding and accurately determining these concentrations, students are equipped to compute the equilibrium constant, which is fundamental to predicting the behavior of the system under different conditions. If these concepts are challenging, students should envision equilibrium concentrations as a 'snapshot' of where the reaction stands when no further changes in concentrations are observed, providing insight into the reaction's dynamics.

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Most popular questions from this chapter

Analysis of a reaction mixture showed that it had the composition \(0.624 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 0.315 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.222 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{NH}_{3}\) at \(800 . \mathrm{K}\), at which temperature \(K_{c}=0.278\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}\) (g). (a) Calculate the reaction quotient \(Q_{c}\). (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

A gaseous mixture consisting of \(1.1 \mathrm{mmol} \mathrm{SO}_{2}\) and \(2.2 \mathrm{mmol}\) \(\mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) container was heated to \(500 . \mathrm{K}\) and allowed to reach equilibrium. Will more sulfur trioxide be formed if that equilibrium mixture is cooled to \(298 \mathrm{~K}\) ? For the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}), K=2.5 \times 10^{10}\) at \(500 . \mathrm{K}\) and \(4.0 \times 10^{24}\) at \(298 \mathrm{~K}\).

(a) Calculate the reaction Gibbs free energy of \(\mathrm{N}_{2}\) (g) \(+\) \(3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) when the partial pressures of \(\mathrm{N}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\) are \(4.2\) bar, \(1.8\) bar, and 21 bar, respectively, and the temperature is 400 . K. For this reaction, \(K=41\) at 400 . K. (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

When \(0.0172 \mathrm{~mol} \mathrm{HI}\) is heated to \(500 . \mathrm{K}\) in a \(2.00-\mathrm{L}\) sealed container, the resulting equilibrium mixture contains \(1.90 \mathrm{~g}\) of HI. Calculate \(K\) for the decomposition reaction \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\)

Write the reaction quotient \(Q\) for (a) \(2 \mathrm{BCl}_{3}(\mathrm{~g})+2 \mathrm{Hg}(\mathrm{l}) \rightarrow \mathrm{B}_{2} \mathrm{Cl}_{4}(\mathrm{~s})+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{~s})\) (b) \(\mathrm{P}_{4} \mathrm{~S}_{10}\) (s) \(+16 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+10 \mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})\) (c) \(\mathrm{Br}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{BrF}_{3}(\mathrm{~g})\)

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