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At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction$$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$.In an experiment, 1.00 mole of \(\mathrm{H}_{2}, 1.00\) mole of \(\mathrm{I}_{2}\), and 1.00 mole of HI are introduced into a 1.00 -L container. Calculate the concentrations of all species when equilibrium is reached.

Short Answer

Expert verified
The equilibrium concentrations for the given reaction are: \([H_2] = 0.7594 M\), \([I_2] = 0.7594 M\), and \([HI] = 1.4812 M\).

Step by step solution

01

Write the given information and set up the ICE table

Initially, we have 1.00 mole of H2, 1.00 mole of I2, and 1.00 mole of HI in a 1.00 L container. First, convert the moles to concentrations by dividing by the volume: [H2] = 1.00 mol / 1.00 L = 1.00 M [I2] = 1.00 mol / 1.00 L = 1.00 M [HI] = 1.00 mol / 1.00 L = 1.00 M Set up the ICE table: ``` Initial Change Equilibrium H2 1.00 M -x 1.00-x I2 1.00 M -x 1.00-x HI 1.00 M +2x 1.00+2x ```
02

Write the equilibrium expression and substitute the equilibrium concentrations

Write the equilibrium expression for K: \(K = \frac{[HI]^2}{[H2][I2]}\) Now, substitute the equilibrium concentrations from the ICE table: \(1.00 \times 10^{2} = \frac{(1.00 + 2x)^2}{(1.00 - x)(1.00 - x)}\)
03

Solve the quadratic equation for x

Multiply both sides of the equation by \((1.00 - x)^2\) and expand: \(100(1.00 - x)^2 = (1.00 + 2x)^2\) Expand the squares: \(100(1 - 2x + x^2) = (1 + 4x + 4x^2)\) Simplify and rearrange to form a quadratic equation: \(96x^2 - 396x + 99 = 0\)
04

Solve for x

We can use the quadratic formula to solve for x: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Using a and b and c from the quadratic equation: \(x = \frac{396 \pm \sqrt{(-396)^2 - 4(96)(99)}}{2(96)}\) We get two possible solutions for x: 0.2406 and 0.4293.
05

Determine the correct value for x and calculate equilibrium concentrations

Check which value of x results in positive concentrations for all species when plugging back into the equilibrium concentrations from the ICE table: For x = 0.2406: [H2] = 1.00 - x = 0.7594 M [I2] = 1.00 - x = 0.7594 M [HI] = 1.00 + 2x = 1.4812 M For x = 0.4293: [H2] = 1.00 - x = 0.5707 M [I2] = 1.00 - x = 0.5707 M [HI] = 1.00 + 2x = 1.8586 M In both cases, we get positive concentrations for all species. However, the value x = 0.4293 would result in a larger change in concentration than the initial moles of reactant, meaning it would fully consume the reactants, which is not possible due to the presence of equilibrium. Therefore, we choose x = 0.2406 as the correct solution. The equilibrium concentrations are: [H2] = 0.7594 M [I2] = 0.7594 M [HI] = 1.4812 M

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Key Concepts

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

Equilibrium Constant (K)
The equilibrium constant, denoted as K, is a fundamental concept in understanding chemical equilibrium. It's a ratio of the concentrations of the products to the reactants at equilibrium, each raised to the power of their stoichiometric coefficients in the balanced equation. It describes the extent to which reactants are converted to products, providing a quantitative measure of the composition of the system at equilibrium.

The value of K is dependent on temperature and is constant for a given reaction at a set temperature. If K is large (much greater than 1), the equilibrium lies towards the products, which indicates that the reaction proceeds almost to completion. Conversely, if K is small (much less than 1), the equilibrium heavily favors the reactants and the reaction proceeds hardly at all. By understanding and calculating the equilibrium constant, we gain insights into the feasibility and directionality of a chemical process.

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) with an equilibrium constant \(K=1.00 \times 10^{2}\), a higher concentration of HI will be favored at equilibrium due to the relatively large value of K.
ICE Table Method
The ICE Table Method is an organized approach to handle calculations involving changes in concentration during reactions. I stands for Initial concentration, C for Change in concentration, and E for Equilibrium concentration. An ICE table helps visualize how concentrations shift from their initial values to new equilibrium values as the reaction proceeds.

This method allows for a systematic way to express the shifts in concentration. For example, a decrease in reactants and an increase in products are denoted by ‘–x’ for reactants and ‘+2x’ for products, where 'x' represents the change. After establishing these relationships, you can then express the equilibrium concentrations in terms of 'x', which can subsequently be used to solve for 'x' using the equilibrium constant expression.

Using the ICE table helps minimize errors and omissions in more complex equilibrium problems by providing a clear and tabular snapshot of the reaction process from start to finish.
Equilibrium Concentration Calculation
The calculation of equilibrium concentrations in a chemical reaction involves finding the concentrations of all reactants and products at the state when the rates of the forward and reverse reactions are equal. This involves using the ICE table in conjunction with the equilibrium constant expression. After setting up the table, we write the expression for K, input the equilibrium concentrations from the ICE table, and solve for 'x'.

Once 'x' is determined, it is substituted back into the expressions for equilibrium concentrations derived from the ICE table. As a vital check, one must ensure that the calculated concentrations are positive and physically meaningful; a negative concentration is not possible and would indicate an error in calculation or assumption.

In our example, 'x' was calculated to be 0.2406, which represents the change from the initial concentration of the reactants (H2 and I2) to equilibrium. This value is used in the ICE table to obtain the final equilibrium concentrations of all species - H2, I2, and HI.
Quadratic Equations in Chemistry
The use of quadratic equations is essential in chemistry when dealing with equilibrium problems that cannot be simplified to linear equations. When the equilibrium constant expression is substituted into the ICE table, often a quadratic equation is formed where the unknown variable 'x' represents the change in concentration of the reactants or products.

To solve a quadratic equation in the form \(ax^2 + bx + c = 0\), we use the quadratic formula \(x = (\text{{-}}b \pm \sqrt{{b^2 - 4ac}}) / (2a)\). Solving this provides two possible values for 'x', and chemical intuition is needed to decide which solution is physically meaningful for the specific reaction context.

For instance, in the solution of our example, we calculated two potential 'x' values using the quadratic formula, but only one made sense chemically as it maintained the integrity of the equilibrium without depleting the reactants completely. Understanding how to solve quadratic equations is thus a crucial skill in the toolkit of any student or professional working with chemical equilibria.

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Most popular questions from this chapter

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned}2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\\P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\\P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm}\end{aligned}$$. Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

In a study of the reaction $$3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g)$$,at \(1200 \mathrm{K}\) it was observed that when the equilibrium partial pressure of water vapor is 15.0 torr, the total pressure at equilibrium is 36.3 torr. Calculate the value of \(K_{\mathrm{p}}\) for this reaction at \(1200 \mathrm{K}\). (Hint: Apply Dalton's law of partial pressures.)

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of \(K\) for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of \(K\) for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.

Suppose a reaction has the equilibrium constant \(K=1.3 \times 10^{8} .\) What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?

In a solution with carbon tetrachloride as the solvent, the compound VCl_ undergoes dimerization:$$2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8}$$ When \(6.6834 \mathrm{g} \mathrm{VCl}_{4}\) is dissolved in \(100.0 \mathrm{g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\). Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(\left.1.696 \mathrm{g} / \mathrm{cm}^{3}, \text { and } K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol} \text { for } \mathrm{CCl}_{4} .\right)\).

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