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

Short Answer

Expert verified
The equilibrium concentrations for the given reaction are approximately: - [H2] = 0.23 M - [I2] = 0.23 M - [HI] = 2.54 M

Step by step solution

01

Write down the equilibrium expression

The given reaction is \[ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \] The equilibrium expression (K) for this reaction will be \[ K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \]
02

Define the initial concentrations and change in concentration

The initial moles of each species are given: - 1 mole of H2 - 1 mole of I2 - 1 mole of HI The volume of the container is 1 L, so we can find the initial concentrations by dividing the moles by the volume. Initial concentrations: - [H2] = 1 M - [I2] = 1 M - [HI] = 1 M As the reaction proceeds, a certain amount (let it be x) of H2 and I2 will react to produce 2x moles of HI. The change in concentration will then be: - [H2] = 1 - x M - [I2] = 1 - x M - [HI] = 1 + 2x M
03

Setup the equation using the equilibrium constant

Given that the equilibrium constant K = 1.00 × 10^2, we can create the following equation by inserting the initial concentrations and the change in concentration into the equilibrium expression: \[ 1.00 \times 10^{2} = \frac{(1 + 2x)^2}{(1 - x)(1 - x)} \]
04

Solve for x

Now, we need to solve this equation to find the value of x. The equation contains a quadratic term, so we can solve it by either factoring or using a quadratic formula. Note: To simplify the process, one can assume the value of x is small, so (1+x) ≈ 1. However, to show the full solution, we will not use this approximation in this case. \[ 100(1 - x)^2 = (1 + 2x)^2 \] Expanding the equation gives: \[ 100 - 200x + 100x^2 = 1 + 4x + 4x^2 \] Rearranging the equation: \[ 96x^2 - 204x + 100 = 0 \] Now, we can use the quadratic formula to find x: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our case, a = 96, b = -204, and c = 100. Plugging these values in: \[ x = \frac{204 \pm \sqrt{(-204)^2 - 4(96)(100)}}{2(96)} \] Solving for x, we get two solutions: x ≈ 1.35 and x ≈ 0.77. Since we cannot have a concentration higher than the initial concentration, we discard the solution x ≈ 1.35 and keep x ≈ 0.77.
05

Calculate the equilibrium concentrations

Now that we have the value of x, we can find the equilibrium concentrations of all species using the expressions from Step 2: - [H2] = 1 - x ≈ 1 - 0.77 = 0.23 M - [I2] = 1 - x ≈ 1 - 0.77 = 0.23 M - [HI] = 1 + 2x ≈ 1 + 2(0.77) = 2.54 M The equilibrium concentrations are approximately: - [H2] = 0.23 M - [I2] = 0.23 M - [HI] = 2.54 M

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, commonly denoted as K, is a crucial concept in chemical equilibrium and reflects the ratio of product concentrations to reactant concentrations, each raised to the power of their respective coefficients in the balanced equation. In the context of the reaction
\(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\),
the equilibrium constant is
\(K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\).
This numerical value is determined at a specific temperature and provides insight into the extent of the reaction; a higher value of K suggests a reaction favoring product formation. When performing calculations, ensure to keep a close eye on the balancing of the reaction since equilibrium constants are sensitive to this balance.
Equilibrium Concentration
Understanding the equilibrium concentration of reactants and products is vital in predicting the outcome of chemical reactions. When the system reaches equilibrium, no further changes in concentrations occur, and the rate of the forward reaction equals the rate of the reverse reaction. In our exercise, the initial concentration of the reactants and products was given as 1 M each. As the reaction progressed, there was a change in the concentrations until equilibrium was established, which is a representation of how dynamic chemical systems can find stability. When calculating equilibrium concentrations, it's essential to determine the change in concentration (
\(x\)) and apply it correctly, as shown by the expressions
\([\mathrm{H}_2] = 1 - x\),
\([\mathrm{I}_2] = 1 - x\),
and
\([\mathrm{HI}] = 1 + 2x\).
This step-by-step approach is a practical guide to quantifying equilibrium states and provides a foundation to explore more complex systems.
Reaction Quotient
The reaction quotient (Q) serves as a predictor of the direction in which a reaction must shift to reach equilibrium. It is calculated in the same manner as the equilibrium constant, using the existing concentrations of reactants and products at a given moment in time. When comparing Q to K, we can determine whether the reaction at hand will proceed forward or reverse to achieve equilibrium. If
\(Q < K\),
the forward reaction is favored, and more products will be formed. Conversely, if
\(Q > K\),
the reverse reaction is favored, thereby increasing the concentration of the reactants. This is a pivotal tool when assessing a reaction's progression toward equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a guiding concept for predicting how a system at equilibrium responds to changes in concentration, temperature, or pressure. It postulates that if any of these conditions are altered, the system will adjust in a manner to counteract the change and re-establish equilibrium. For example, if additional reactant is introduced to a system, the reaction will shift towards the products to reduce the impact of this change. Conversely, increasing the product concentration would shift the reaction towards the reactants. This principle aids in understanding and controlling chemical reactions, ensuring that predictions and adjustments can be made under various conditions to reach desired chemical states.

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Most popular questions from this chapter

The equilibrium constant \(K_{\mathrm{p}}\) for the reaction $$ \mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) $$ at \(700^{\circ} \mathrm{C}\) is \(0.76 .\) Detemine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of \(1.20 \mathrm{~atm}\) at \(700^{\circ} \mathrm{C}\).

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.

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is \(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \quad K=5.6 \times 10^{-11}\) If \(0.16\) mole of \(\mathrm{HCO}_{3}-\) is placed into \(1.00 \mathrm{~L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-}\) ?

For the following endothermic reaction at equilibrium: $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ which of the following changes will increase the value of \(K ?\) a. increasing the temperature b. decreasing the temperature c. removing \(\mathrm{SO}_{3}(g)\) (constant \(T\) ) d. decreasing the volume (constant \(T\) ) e. adding \(\operatorname{Ne}(g)\) (constant \(T\) ) f. adding \(\mathrm{SO}_{2}(g)\) (constant \(T\) ) g. adding a catalyst (constant \(T\) )

Consider the decomposition equilibrium for dinitrogen pentoxide: $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ At a certain temperature and a total pressure of \(1.00 \mathrm{~atm}\), the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0\), will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to \(0.50 \%\) ? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of \(10.0 .\)

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