Using Le Chatelier's Principle with a change of concentration Suppose you have an equilibrium established between four substances A, B, C and D. Apart from using the equilibrium constant for predicting the direction of the reaction, the value of the reaction quotient also plays a crucial role. when a catalyst is added to a reaction at equilibrium. An astute gardener can adjust the pH of the soil and actually change the color of the flowers. This situation is represented in Figure \(\PageIndex{3}\), which shows a plot of \([\ce{CO_2}]\) versus the amount of \(\ce{PbCO_3}\) added. { "6.01:_Reversible_Reaction" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.02:_Chemical_Equilibrium" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.03:_Equilibrium_Constant" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.04:_Interpreting_Equilibrium_Constants_Values" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.05:_Calculations_involving_Equilibrium_Constants_and_Concentrations_at_Equilibrium" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.06:_Calculations_involving_Equilibrium_Constants_and_Initial_Concentrations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.07:_Predicting_the_Direction_of_a_Reaction-_the_reaction_quotient" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.08:_Shifting_Equilibria_-_Le_Chatelier\'s_Principle" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.09:_Relationship_Between_Free_Energy_and_(K_texteq)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "6.10:_Nonreversible_Reactions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "01:_Electrochemistry" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "02:_The_States_of_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "03:_Solutions_and_Colloids" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "04:_Thermochemistry_and_Thermodynamics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "05:_Reaction_Rates" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "06:_Chemical_Equilibrium" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "07:_Acid_and_Base_Equilibria" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "08:_Radioactivity_and_Nuclear_Processes" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()" }, 6.8: Shifting Equilibria - Le Chatelier's Principle, [ "article:topic", "catalyst", "equilibrium", "Le Chatelier\'s Principle", "showtoc:no", "Keq", "Le Ch\u00e2telier\u2019s principle", "license:ccbyncsa", "source[1]-chem-64089", "source[2]-chem-64089", "licenseversion:40" ], https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2FBrevard_College%2FCHE_104%253A_Principles_of_Chemistry_II%2F06%253A_Chemical_Equilibrium%2F6.08%253A_Shifting_Equilibria_-_Le_Chatelier's_Principle, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), increased pressure, decreased temperature, removal of NH, 6.7: Predicting the Direction of a Reaction- the reaction quotient, 6.9: Relationship Between Free Energy and \(K_\text{eq}\), Effect of changing concentrations of reactants or products. Solved Which direction would you predict the equilibrium for - Chegg Both are values of the ratio of the concentrations of the products to the reactants. pressures stay the same. For example, we have 1.2 10 2mol of CH 4 in a 2.0 L container, so. By knowing the values of the equilibrium constant and the reaction quotient of a chemical reaction, the direction of the reaction can be predicted. {eq}\hspace{2cm} {/eq}This is a weak acid; therefore, the equilibrium reaction is {eq}HF(aq) + H_2O(l) \rightleftharpoons H_3O^+ + F^- {/eq}. b. Which direction would you predict the equilibrium for each reaction to be shifted? For AQA GCSE Chemistry, the specific details of how ammonia is. The reaction is in the backward direction when the chemical reaction quotient is greater than the equilibrium constant. Download our apps to start learning, Call us and we will answer all your questions about learning on Unacademy. We need a term, reaction quotient (Qc expressed in terms of concentrations or Qp in terms of partial pressures) similar to the equilibrium constant, except that the conditions are not at equilibrium. The quantities involved for each substance are their thermodynamic activities, which are approximated by partial pressures for gases, molar concentrations for solutions , and 1 for pure liquids and solids. Ans. So there's no shift In more basic soils, aluminum is less soluble, and under these conditions the hydrangea flowers are red. Test Yourself. Again, only those pairs of concentrations of \(\ce{H_2O}\) and \(\ce{H_2}\) that lie on the line correspond to equilibrium states. lessons in math, English, science, history, and more. - Definition & History, Temperature Units: Converting Between Kelvin and Celsius, 7th Grade Louisiana Social Studies State Standards, 8th Grade Louisiana Social Studies State Standards, 6th Grade Louisiana Social Studies State Standards, Alabama Foundations of Reading (190): Study Guide & Prep, How to Apply for College Grants & Scholarships. Each experiment begins with different proportions of product and reactant: As these calculations demonstrate, \(Q\) can have any numerical value between 0 and infinity (undefined); that is, \(Q\) can be greater than, less than, or equal to \(K\). ET Sunday, the National Hurricane Center said the . In which directiontoward reactants or toward productsdoes the reaction shift if the equilibrium is stressed by each change? From food preparation and digestion to fuel formation, chemical reactions play a crucial role. When the value of Qc is lesser than the value of Kc, then. Poison Dart Frog | Characteristics, Habitat & Facts, John Muir: Facts, Books & Accomplishments, Julio Cortazar: Biography, Short Stories & Poems. The reaction Quotient (\(Q\)) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction. Predicting the Direction of Reaction If Qc > Kc, the reaction will shift left; toward reactants If Qc < Kc, the reaction will shift right; toward products If Qc = Kc, then the reaction is at equilibrium as follows: Reactants Products Consider the following equilibrium: N 2 (g) + 3H 2 (g) 2NH 3 (g) [CH 4] = 1.2 10 2mol 2.0 L = 6.0 10 3M. Next, the volume is increased on the reaction at equilibrium. It is bidirectional, happening in both directions. When we stress the equilibrium, the chemical reaction is no longer at equilibrium, and the reaction starts to move back toward equilibrium in such a way as to decrease the stress. (Select ] shifts in forward direction no shift Add cd shifts in reverse direction. Interestingly, the color of the flowers is due to the acidity of the soil that the hydrangea is planted in. When more reactants are added to the reaction mixture, the direction of the reaction shifts from left to right, reactants form the products efficiently. We also learn the importance of XeF6 molecular geometry and bond angles importance and much more about the topic in detail. Changing the position of equilibrium - Higher - Reversible - BBC To reach equilibrium, the system must decrease \([\ce{CO_2}]\), which it can do only by reacting \(\ce{CO_2}\) with solid \(\ce{PbO}\) to form solid \(\ce{PbCO_3}\). When both the values are equal, the reaction halts. Comparing the magnitudes of \(Q\) and \(K\) enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). Thus the left portion of the graph represents a system that is not at equilibrium because it contains only \(\ce{CO2(g)}\) and \(\ce{PbO(s)}\). Althoughthere are several ways to stress an equilibrium, as we will see below. If the temperature of the system is decreased, the reverse effect will be . Step 3: Compare it to K. If K > Q, the reaction goes spontaneously forward. By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which \(Q = K\). The only difference is while calculating the reaction quotient, the reaction is not at equilibrium. The little arrow points to the left or toward the reverse direction of the reaction. Human Chorionic Gonadotropin (HCG): Side Effects & Carotenoid Pigments: Definition & Structure. For example, in the following example. {eq}\hspace{2cm} K > Q {/eq} and the reaction goes forward. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss. Le Chatelier's Principle in Chemistry - ThoughtCo However, a catalyst does not affect the extent or position of a reaction at equilibrium. To determine whether a system has reached equilibrium, chemists use a Quantity called the reaction Quotient (\(Q\)). For example, if the temperature is increased for an endothermic reaction, essentially a reactant is being added, so the equilibrium shifts toward products. Use the Equilibrium constant for predicting the direction of a reaction, IIT JEE notes. \[PCl_{3}+Cl_{2}\rightleftharpoons PCl_{5}+60kJ\nonumber \]. According to the Le Chatelier's principle, the net reaction will Ans. \[2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g)+196kJ\nonumber \], Given this equilibrium, predict the direction of shift for each stress. If \(Q = K\), then the system is at equilibrium. Predict the direction of shift for an equilibrium under stress. For example, point A in Figure \(\PageIndex{3}\) lies below the line, indicating that the \([\ce{H_2O}]/[\ce{H_2}]\) ratio is less than the ratio of an equilibrium mixture (i.e., \(Q < K\)). Both are values of the ratio of the concentrations of the products to the reactants. will react with hydrogen gas to produce methanol. It is worth noting that when reactants or products are added or removed, the value of the Keq does not change. Once equilibrium is established, the reaction is over, right? blue b. heat + Co2+ (aq) + 4C1 (aq) = CoCl22- (aq); Co2+ is pink colorless . which direction that is by looking at the balanced equation. This stage in a reversible reaction is where the concentration of the reactants is equal to that of the products, according to the law of mass action. However, the catalyst is gonna equilibrium will shift, to the left, to the right, or not at all, as we try to make changes to Here are two example problems to help you use an equilibrium constant to predict the direction of spontaneous reaction. Physical Science for Teachers: Professional Development, Anatomy & Physiology for Teachers: Professional Development. For the reaction H2+I22HI, consider two possibilities: (a) you add 0.5 mole of each reactant, allow the system to come to equilibrium, and then add 1 mole of H2, and allow the system to reach equilibrium again, or (b) you add 1.5 moles of H2and 0.5 mole of 12 and allow the system to come to equilibrium. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In the watergas shift reaction introduced in Example \(\PageIndex{1}\), carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen: \[\ce{CO(g) + H_2O(g) <=> CO2(g) + H2(g)} \nonumber \]. Whereas while calculating the equilibrium constant, the reaction is at equilibrium. Term 1 / 16 Le Chtelier's principle Click the card to flip Definition 1 / 16 Le Chtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. Unacademy is Indias largest online learning platform. In such cases, the reaction in Equation \(\ref{15.6.4}\) will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. 6.8: Shifting Equilibria - Le Chatelier's Principle If the temperature of the system is increased (at constant V), the system will shift in the direction that consumes the excess heat. What is the effect on this equilibrium if pressure is decreased? for an exothermic reaction. pKa = -log Ka The lower the pKa value of an acid, the stronger the acid. Calculate the molar concentrations of the reactants and the products. Solving Problems Involving Systems of Equations, Agriculture & Irrigation of the Chola Dynasty, Theodicy Philosophy: Definition & Overview. An equilibrium constant is the ratio of the concentration of the products to that of the reactants. Created by Yuki Jung. And that might sound a When a reaction system is at equilibrium, \(Q = K\). and two moles of gas for a total of three moles of gas. A reaction quotient is the same as the equilibrium constant, except the reaction is not at equilibrium. Once equilibrium is established, the reaction is over, right? Thus, the direction of the reaction can be predicted from Vapour density measurements. Say when I start or initially I have one molar of my reactant and I have none of by-product B. Use Equation \(\ref{15.6.1}\) to determine \(Q\). It only takes a few minutes. The expression for the reaction Quotient has precisely the same form as the equilibrium constant expression, except that \(Q\) may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and the products, regardless of whether the system is at equilibrium. The Reaction Quotient. It depends on whether the reaction is endothermic or exothermic. We previously saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Additionally, he holds master's degrees in chemistry and physician assistant studies from Villanova University and the University of Saint Francis, respectively. This process is described by Le Chtelier's principle: When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. How can you predict the direction of equilibrium? Why change of temperature would change equilibrium constant but if we changed the volume the constant still remained the same? Thus the reaction in Equation \(\ref{15.6.4}\) will proceed to the left as written, consuming \(\ce{H_2O}\) and producing \(\ce{H_2}\), which causes the concentration ratio to move down and to the right toward the equilibrium line. What is the effect of temperature changes on an equilibrium? The reaction quotient Q (article) | Khan Academy The only difference is while calculating the reaction quotient, the reaction is not at equilibrium. However, pressure strongly impacts the gas phase. The reaction in Equation \ref{15.6.3} will therefore proceed to the right as written, until \([\ce{CO_2}] = K\). For example, in the {eq}NO_2 {/eq} reaction. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. All rights reserved. {eq}\hspace{2cm}Q = \dfrac{\left[H_3O^+\right]_0\left[F^-\right]_0}{\left[HF\right]_0}\hspace{1cm} {/eq}Adding sodium fluoride increases {eq}\left[F^-\right]_0 {/eq}; therefore, Q gets larger. in the following equilibria when the indicated stress is applied: a. heat + Co2+ (aq) + 4C1 (aq) = CoC14?- (aq); The pink colorless equilibrium mixture is heated. shift to increase the pressure. Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Predicting The Direction Of A Reaction - BYJU'S Since the net reaction is going to try to increase the pressure, the equilibrium shifts to the left, toward the side that's gonna In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor: \[\ce{CdO(s) + H2(g) <=> Cd(s) + H_2O(g)} \label{15.6.4} \], \[K = \dfrac{[\ce{H_2O}]}{[\ce{H_2}]}. )%2F15%253A_Principles_of_Chemical_Equilibrium%2F15.5%253A_The_Reaction_Quotient_Q_-_Predicting_The_Direction_of_Net_Change, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), 15.4: The Magnitude of an Equilibrium Constant, 15.6: Altering Equilibrium Conditions - Le Chteliers Principle, Predicting the Direction of a Reaction with a Graph, Using the Reaction Quotient (Q) (opens in new window), \(\dfrac{(0.0600)^2}{0}=\text{undefined}\), \(\dfrac{(0.0200)^2}{0.0600}=6.67 \times 10^{3}\). At elevated temperatures, methane (\(CH_4\)) reacts with water to produce hydrogen and carbon monoxide in what is known as a steam-reforming reaction: \[\ce{CH4(g) + H2O(g) <=> CO(g) + 3H2(g)} \nonumber \]. What is a Covenant of Seisin? Get answers to the most common queries related to the JEE Examination Preparation. Use Equation \(\ref{15.6.1}\) to determine \(Q\). If the number of moles of gas is the same on both sides of the reaction, pressure has no effect. {eq}\hspace{2cm}Q = \dfrac{(p_{H_{2}})_0^4}{(p_{H_{2}O})_0^4}\hspace{1cm} {/eq}If K = Q at first, then adding more steam will make {eq}(p_{H_{2}O})_0 {/eq} bigger and Q smaller. Learning Objectives To predict in which direction a reaction will proceed. a) Shift to b) shift to c) No. Given: balanced chemical equation, \(K\), amounts of reactants and products, and volume, A We must first find the initial concentrations of the substances present. {eq}\hspace{2cm} {/eq}The reaction is written out for you. A general overview of Lewis Structure, XeF4 Molecular Geometry and bond Angles meaning, valuable XeF4 Molecular Geometry and bond angle questions. heat on the product side. Suppose you have two identical flasks, connected by a stopcock. Accessibility StatementFor more information contact us atinfo@libretexts.org. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination. little strange at first because adding neon gas means that the total pressure would increase, the total pressure since Conversely, any point that lies above and to the right of the equilibrium curve (such as point B in Figure \(\ref{15.6.5}\)) corresponds to \(Q > K\), and the reaction in Equation \(\ref{15.6.5}\) will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. The left side contains the reactants. This video explains the differences between reaction quotient (Q) and equilibrium constant (K), and shows an example of how to compare Q and K to determine t. This video explains the. Such a graph allows us to predict what will happen to a reaction when conditions change so that \(Q\) no longer equals \(K\), such as when a reactant concentration or a product concentration is increased or decreased. Not exactly. Let's say that the C + O2 CO2 [Carbon + Oxygen Carbon dioxide], N2 + O2 2NO [Nitrogen + Oxygen 2 Nitrous Oxide]. Thanks in advance. Conversely, the point labeled B in Figure \(\PageIndex{2}\) lies below the horizontal line, so it corresponds to a \([\ce{CO_2}]\) that is less than the equilibrium concentration of \(\ce{CO_2}\) (i.e., \(Q < K\)). \([\ce{CO}] = 8.0 \times 10^{3} M\), and. Le Chatelier's principle implies that a pressure increase shifts an equilibrium to the side of the reaction with the fewer number of moles of gas, while a pressure decrease shifts an equilibrium to the side of the reaction with the greater number of moles of gas. In other words the reaction will go in reverse direction until equilibrium is reached. An experimenter has some ability to affect the equilibrium. The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of \(Q\) were calculated for each. Compare Q and K to determine in which direction the reaction will proceed. For example, the point labeled A in Figure \(\PageIndex{2}\) lies above the horizontal line, so it corresponds to a \([\ce{CO_2}]\) that is greater than the equilibrium concentration of \(\ce{CO_2}\) (i.e., \(Q > K\)). Thus, adding more products during a forward reaction shifts it towards the right to the left. The reaction will therefore proceed to the right as written, forming \(\ce{H2}\) and \(\ce{CO}\) at the expense of \(\ce{H_2O}\) and \(\ce{CH4}\). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes \(Q\) approach \(K\): These points are illustrated graphically in Figure \(\PageIndex{1}\). the reaction quotient is defined as follows: \[Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{15.6.1} \]. Therefore, for the following general reaction: \[aA+bB \rightleftharpoons cC+dD \nonumber \]. A plot of \([\ce{H_2O}]\) versus \([\ce{H_2}]\) at equilibrium is a straight line with a slope of \(K\) (Figure \(\PageIndex{3}\)). Therefore, the reaction will proceed to the, If \(Q > K\), then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium, so the reaction will proceed to the. Because \(K = 2.4 \times 10^{4}\), we see that \(Q < K\). 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https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FCourses%2Fcan%2Fgeneral%2F16%253A_Chemical_Equilibrium%2F16.06%253A_The_Reaction_Quotient-_Predicting_the_Direction_of_Change, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), 16.5: Heterogenous Equilibria - Reactions Involving Solids and Liquids, Predicting the Direction of a Reaction with a Graph, Using the Reaction Quotient (Q) (opens in new window), \(\dfrac{(0.0600)^2}{0}=\text{undefined}\), \(\dfrac{(0.0200)^2}{0.0600}=6.67 \times 10^{3}\).
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