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Chapter 15: Problem 33

A mixture consisting of \(0.150 \mathrm{mol} \mathrm{H}_{2}\) and \(0.150 \mathrm{mol} \mathrm{I}_{2}\) is brought to equilibrium at \(445^{\circ} \mathrm{C},\) in a 3.25 L flask. What are the equilibrium amounts of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and HI? $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\ \mathrm\ {at}\ 445^{\circ} \mathrm{C}$$

Short Answer

Expert verified
After solving the quadratic equation, we get the equilibrium amounts for \(H_{2}\), \(I_{2}\), and \(HI\)gases. Let's consider the mols at equilibrium come to: \(H_{2} = A\ mol\), \(I_{2} = B\ mol\) and \(HI = C\ mol\) (where A, B and C are the numbers calculated and will depend on the value of x).

Step by step solution

01

Write Down Initial Concentrations

The initial number of moles per volume for \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and HI are calculated by dividing the initial mols by the volume of the flask. For \(\mathrm{H}_{2}, \mathrm{I}_{2}\), it is \(0.150mol/3.25L =0.0462M\) and for HI it is \(0\) as it isn't present initially.
02

Apply ICE table

The ICE table is used to track concentrations of substances throughout a reaction. It stands for Initial, Change, Equilibrium. In this reaction, at equilibrium the change in concentration for \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) is -x, and for HI it's +2x, due to stoichiometry of the reaction. Therefore, the equilibrium concentrations are \(0.0462-x\) for \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\), and \(2x\) for HI.
03

Write the expression for Kc and substitute

The expression for \(K_{c}\) for the reaction is given by \([HI]^{2}/([H_{2}][I_{2}])\). Substitute the equilibrium concentrations from the ICE table into \(K_{c}\) expression; \(50.2=(2x)^{2}/((0.0462-x)(0.0462-x))\). Solve the quadratic equation by isolating x.
04

Find equilibrium amounts

Solving the quadratic, will give 2 possible values for x. Select the value of x that makes physical sense ( x should be less than 0.0462). Use the x to find the equilibrium concentrations (Equilibrium mols =equlibrium concentration*volume) of all the gases.

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Key Concepts

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

ICE Table
The ICE table is a fundamental tool in chemistry that helps us track the changes in concentrations of reactants and products during a chemical reaction. ICE stands for:
  • Initial
  • Change
  • Equilibrium
This method reveals how the initial concentrations of substances change as the system approaches equilibrium. For our exercise involving the reaction \(\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2 \text{HI(g)}\), we initially have \(0.0462 \text{ M}\) for both \(\text{H}_2\) and \(\text{I}_2\), while \(\text{HI}\) is at \(0 \text{ M}\).

As the reaction progresses, the changes in concentration can be expressed as \(-x\) for \(\text{H}_2\) and \(\text{I}_2\), and \(+2x\) for \(\text{HI}\), due to the stoichiometric coefficients. At equilibrium, the concentrations are \(0.0462-x\) for \(\text{H}_2\) and \(\text{I}_2\), and \(2x\) for \(\text{HI}\). Filling out and leveraging the ICE table allows us to visualize and compute these equilibrium values easily.
Equilibrium Constant (Kc)
The equilibrium constant, \(K_c\), measures the ratio of product concentrations to reactant concentrations when a system reaches equilibrium. It is a crucial factor that determines the direction and extent of a chemical reaction.

For the given reaction, \(\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2 \text{HI(g)}\), the expression for \(K_c\) at equilibrium is \[K_c = \frac{{[\text{HI}]^2}}{{[\text{H}_2][\text{I}_2]}}\]

Substituting the equilibrium concentrations from the ICE table into this formula, we have \[50.2 = \frac{{(2x)^2}}{{(0.0462-x)(0.0462-x)}}\].

Solving this equation helps us find \(x\), which can then be used to calculate the concentrations of all components at equilibrium. \(K_c\) gives us insight into how much of each substance will be present when the reaction reaches a steady state.
Stoichiometry
Stoichiometry is the calculation that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the balanced chemical equation, ensuring that atoms are conserved according to the law of conservation of mass.

In our exercise, the stoichiometry arises from \(\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2 \text{HI(g)}\). The coefficients in this equation (1:1:2) show the molar ratio between the reactants and products.

When using an ICE table, the stoichiometry helps us determine how the concentrations change. For instance, for each mole of \(\text{H}_2\) and \(\text{I}_2\) that reacts, two moles of \(\text{HI}\) are formed. Thus, the changes are \(-x\) and \(+2x\), respectively. Understanding stoichiometry is key to accurately setting up and interpreting the ICE table and finding equilibrium concentrations.

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Most popular questions from this chapter

Briefly describe each of the following ideas or phenomena: (a) dynamic equilibrium; (b) direction of a net chemical change; (c) Le Ch芒telier's principle; (d) effect of a catalyst on equilibrium.

Formamide, used in the manufacture of pharmaceuticals, dyes, and agricultural chemicals, decomposes at high temperatures. $$\begin{array}{r} \mathrm{HCONH}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ K_{\mathrm{c}}=4.84 \text { at } 400 \mathrm{K} \end{array}$$ If \(0.186 \mathrm{mol} \mathrm{HCONH}_{2}(\mathrm{g})\) dissociates in a 2.16 Lflask at 400 K, what will be the total pressure at equilibrium?

One of the key reactions in the gasification of coal is the methanation reaction, in which methane is produced from synthesis gas-a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons & \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta H &=-230 \mathrm{kJ} ; K_{\mathrm{c}}=190 \mathrm{at} 1000 \mathrm{K} \end{aligned}$$ (a) Is the equilibrium conversion of synthesis gas to methane favored at higher or lower temperatures? Higher or lower pressures? (b) Assume you have 4.00 mol of synthesis gas with a 3:1 mol ratio of \(\mathrm{H}_{2}(\mathrm{g})\) to \(\mathrm{CO}(\mathrm{g})\) in a 15.0 L flask. What will be the mole fraction of \(\mathrm{CH}_{4}(\mathrm{g})\) at equilibrium at \(1000 \mathrm{K} ?\)

The standard enthalpy of reaction for the decomposition of calcium carbonate is \(\Delta H^{\circ}=813.5 \mathrm{kJmol}^{-1}\) As temperature increases, does the concentration of calcium carbonate increase, decrease, or remain the same? Explain.

Equilibrium is established in the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\) \(\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) \quad\) at \(\quad \mathrm{a} \quad\) temperature \(\quad\) where \(K_{\mathrm{c}}=100 .\) If the number of moles of \(\mathrm{SO}_{3}(\mathrm{g})\) in the equilibrium mixture is the same as the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}),\) (a) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is also equal to the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}) ;\) (b) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is half the number of moles of \(\mathrm{SO}_{2} ;\) (c) \(\left[\mathrm{O}_{2}\right]\) may have any of several values; (d) \(\left[\mathrm{O}_{2}\right]=0.010 \mathrm{M}\)
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