Only the answer with the positive value has any physical significance, so \([H_2O] = [CO] = +0.148 M\), and \([H_2] = [CO_2] = 0.148\; M\). It is important to remember that even though the concentrations are constant at equilibrium, the reaction is still happening! The reaction is already at equilibrium! Initial reactant and product concentrations and equilibrium concentrations (in M) are given as well as the equilibrium constants (at 25 C). The equilibrium mixture contained. In other words, the concentration of the reactants is higher than it would be at equilibrium; you can also think of it as the product concentration being too low. Write the equilibrium constant expression for each reaction. with \(K_p = 2.5 \times 10^{59}\) at 25C. Direct link to Eun Ju Jeong's post You use the 5% rule when , Posted 7 years ago. Solved Part A) At equilibrium the reactant and product - Chegg Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. The final \(K_p\) agrees with the value given at the beginning of this example. Under certain conditions, oxygen will react to form ozone, as shown in the following equation: \[3O_{2(g)} \rightleftharpoons 2O_{3(g)}\nonumber \]. Example 15.7.1 If you're seeing this message, it means we're having trouble loading external resources on our website. What is the partial pressure of NO in equilibrium with \(N_2\) and \(O_2\) in the atmosphere (at 1 atm, \(P_{N_2} = 0.78\; atm\) and \(P_{O_2} = 0.21\; atm\)? why shouldn't K or Q contain pure liquids or pure solids? The equilibrium constant of a chemical reaction is the value of the reaction quotient when the reaction has reached equilibrium. By looking at the eq position you can determine if the reactants or products are favored at equilibrium Reactant>product reaction favors reactant side Product>reactant reaction favors product side - Eq position is largely determind by the activation energy of the reaction If . The initial concentrations of \(NO\) and \(Cl_2\) are \(0\; M\) because initially no products are present. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of \(x\). Say if I had H2O (g) as either the product or reactant. Direct link to Brian Walsh's post I'm confused with the dif, Posted 7 years ago. Direct link to Chris's post http://www.chem.purdue.ed, Posted 7 years ago. C Substituting this value of \(x\) into our expressions for the final partial pressures of the substances. If the equilibrium favors the products, does this mean that equation moves in a forward motion? For reactions that are not at equilibrium, we can write a similar expression called the. Only in the gaseous state (boiling point 21.7 C) does the position of equilibrium move towards nitrogen dioxide, with the reaction moving further right as the temperature increases. If we define \(x\) as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is \(+x\). What is \(K\) for the reaction, \[N_2+3H_2 \rightleftharpoons 2NH_3\nonumber \], \(K = 0.105\) and \(K_p = 2.61 \times 10^{-5}\), A Video Disucssing Using ICE Tables to find Kc: Using ICE Tables to find Kc(opens in new window) [youtu.be]. Conversion of K c to K p To convert K c to K p, the following equation is used: Kp = Kc(RT)ngas where: R=0.0820575 L atm mol -1 K -1 or 8.31447 J mol -1 K -1 Example \(\PageIndex{3}\) illustrates a common type of equilibrium problem that you are likely to encounter. The equilibrium constant expression would be: which is the reciprocal of the autoionization constant of water (\(K_w\)), \[ K_c = \dfrac{1}{K_w}=1 \times 10^{14}\]. The yellowish sand is covered with people on beach towels, and there are also some swimmers in the blue-green ocean. Can i get help on how to do the table method when finding the equilibrium constant. When we aren't sure if our reaction is at equilibrium, we can calculate the reaction quotient, At this point, you might be wondering why this equation looks so familiar and how. A Because we are given Kp and partial pressures are reported in atmospheres, we will use partial pressures. Some will be PDF formats that you can download and print out to do more. The equilibrium mixture contained. This \(K\) value agrees with our initial value at the beginning of the example. The equilibrium constant expression is an equation that we can use to solve for K or for the concentration of a reactant or product. At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. The concentration of dinitrogen tetroxide starts at an arbitrary initial concentration, then decreases until it reaches the equilibrium concentration. Using the Haber process as an example: N 2 (g) + 3H 2 (g . Which of the following happens when a reaction reaches - Brainly To obtain the concentrations of \(NOCl\), \(NO\), and \(Cl_2\) at equilibrium, we construct a table showing what is known and what needs to be calculated. The problem then is identical to that in Example \(\PageIndex{5}\). Check out 'Buffers, Titrations, and Solubility Equilibria'. Write the equilibrium constant expression for the reaction. Otherwise, we must use the quadratic formula or some other approach. the rates of the forward and reverse reactions are equal. Substitute appropriate values from the ICE table to obtain \(x\). Atmospheric nitrogen and oxygen react to form nitric oxide: \[N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber \]. When can we make such an assumption? As the reaction proceeds, the concentrations of CO . The results we have obtained agree with the general observation that toxic \(NO\), an ingredient of smog, does not form from atmospheric concentrations of \(N_2\) and \(O_2\) to a substantial degree at 25C. the rates of the forward and reverse reactions are equal. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations). Taking the square root of the middle and right terms, \[\dfrac{x}{(0.0150x)} =(0.106)^{1/2}=0.326\nonumber \], \[x =0.00369=3.69 \times 10^{3}\nonumber \]. The Equilibrium Constant is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. Thus, the units are canceled and \(K\) becomes unitless. Direct link to abhishekppatil99's post If Kc is larger than 1 it, Posted 6 years ago. Most of these cases involve reactions for which the equilibrium constant is either very small (\(K 10^{3}\)) or very large (\(K 10^3\)), which means that the change in the concentration (defined as \(x\)) is essentially negligible compared with the initial concentration of a substance. if the reaction will shift to the right, then the reactants are -x and the products are +x. Construct a table showing the initial concentrations of all substances in the mixture. All reactions tend towards a state of chemical equilibrium, the point at which both the forward process and the reverse process are taking place, Based on the concentrations of all the different reaction species at equilibrium, we can define a quantity called the equilibrium constant. Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations. In a chemical reaction, when both the reactants and the products are in a concentration which does not change with time any more, it is said to be in a state of chemical equilibrium. The new expression would be written as: \[K'= \dfrac{1}{\dfrac{[G]^g[H]^h}{[A]^a[B]^b}} = \dfrac{[A]^a[B]^b}{[G]^g[H]^h}\]. PDF Chapter 15 Chemical Equilibrium - University of Pennsylvania Direct link to Alejandro Puerta-Alvarado's post I get that the equilibr, Posted 5 years ago. You use the 5% rule when using an ice table. Then substitute values from the table into the expression to solve for \(x\) (the change in concentration). Activity is expressed by the dimensionless ratio \(\frac{[X]}{c^{\circ}}\) where \([X]\) signifies the molarity of the molecule and c is the chosen reference state: For gases that do not follow the ideal gas laws, using activities will accurately determine the equilibrium constant that changes when concentration or pressure varies. A K of any value describes the equilibrium state, and concentrations can still be unchanging even if K=!1. Direct link to S Chung's post Check out 'Buffers, Titra, Posted 7 years ago. Calculate the partial pressure of \(NO\). The equilibrium constant of pressure gives the ratio of pressure of products over reactants for a reaction that is at equilibrium (again, the pressures of all species are raised to the powers of their respective coefficients). Direct link to Lily Martin's post why aren't pure liquids a, Posted 6 years ago. Calculate the final concentrations of all species present. There are two fundamental kinds of equilibrium problems: We saw in the exercise in Example 6 in Section 15.2 that the equilibrium constant for the decomposition of \(CaCO_{3(s)}\) to \(CaO_{(s)}\) and \(CO_{2(g)}\) is \(K = [CO_2]\). Direct link to Ibeh JohnMark Somtochukwu's post the reaction quotient is , Posted 7 years ago. Hooray! The contents of the reactor were then analyzed and found to contain 0.056 mol of \(Cl_2\). Direct link to Cynthia Shi's post If the equilibrium favors, Posted 7 years ago. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions. The initial partial pressure of \(O_2\) is 0.21 atm and that of \(N_2\) is 0.78 atm. A ratio of molarities of products over reactants is usually used when most of the species involved are dissolved in water. Direct link to Emily's post YES! From a mathematical perspective, with the activities of solids and liquids and solvents equal one, these substances do not affect the overall K or Q value. . Given: balanced equilibrium equation and composition of equilibrium mixture. Check your answer by substituting values into the equilibrium equation and solving for \(K\). A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. The equilibrium constant K (article) | Khan Academy The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. In this state, the rate of forward reaction is same as the rate of backward reaction. For the same reaction, the differing concentrations: \[SO_{2 (g)} = 0.1\; M O_{2(g)} = 0.3\; M \;SO_{3 (g)} = 0.5\; M\] Would this go towards to product or reactant? K is the equilibrium constant. Thus the equilibrium constant for the reaction as written is 2.6. Or would it be backward in order to balance the equation back to an equilibrium state? Example 10.3.4 Determine the value of K for the reaction SO 2(g) + NO 2(g) SO 3(g) + NO(g) when the equilibrium concentrations are: [SO 2] = 1.20M, [NO 2] = 0.60M, [NO] = 1.6M, and [SO 3] = 2.2M. \(P_{NO}=2x \; atm=1.8 \times 10^{16} \;atm \). , Posted 7 years ago. Thus we must expand the expression and multiply both sides by the denominator: \[x^2 = 0.106(0.360 1.202x + x^2)\nonumber \]. Insert those concentration changes in the table. Chemistry Chapter 13: Equilibrium Concepts Study Guide We enter the values in the following table and calculate the final concentrations. When a chemical system is at equilibrium, A. the concentrations of the reactants are equal to the concentrations of the products B the concentrations of the reactants and products have reached constant values C. the forward and reverse reactions have stopped. Check your answers by substituting these values into the equilibrium constant expression to obtain \(K\). In Example \(\PageIndex{3}\), the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. What is the composition of the reaction mixture at equilibrium? Again, \(x\) is defined as the change in the concentration of \(H_2O\): \([H_2O] = +x\). B We can now use the equilibrium equation and the known \(K\) value to solve for \(x\): \[K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{x^2}{(0.570x)(0.632x)}=0.106\nonumber \]. Keyword- concentration. \([H_2]_f[ = [H_2]_i+[H_2]=0.570 \;M 0.148\; M=0.422 M\), \([CO_2]_f =[CO_2]_i+[CO_2]=0.632 \;M0.148 \;M=0.484 M\), \([H_2O]_f =[H_2O]_i+[H_2O]=0\; M+0.148\; M =0.148\; M\), \([CO]_f=[CO]_i+[CO]=0 M+0.148\;M=0.148 M\). "Kc is often written without units, depending on the textbook.". Direct link to Ernest Zinck's post As you say, it's a matter, Posted 7 years ago. As a general rule, if \(x\) is less than about 5% of the total, or \(10^{3} > K > 10^3\), then the assumption is justified. 13.1 Chemical Equilibria - Chemistry 2e | OpenStax in the example shown, I'm a little confused as to how the 15M from the products was calculated. Direct link to Everett Ziegenfuss's post Would adding excess react, Posted 7 years ago. The same process is employed whether calculating \(Q_c\) or \(Q_p\). 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. The equilibrium constant for this reaction is 0.030 at 250 o C. Assuming that the initial concentration of PCl 5 is 0.100 moles per liter and there is no PCl 3 or Cl 2 in the system when we start, let's calculate the concentrations of PCl 5, PCl 3, and Cl 2 at equilibrium. If the product of the reaction is a solvent, the numerator equals one, which is illustrated in the following reaction: \[ H^+_{(aq)} + OH^_{(aq)} \rightarrow H_2O_{ (l)}\]. In reaction B, the process begins with only HI and no H 2 or I 2. Direct link to Amrit Madugundu's post How can we identify produ, Posted 7 years ago. By comparing. Write the equilibrium equation for the reaction. Substituting the appropriate equilibrium concentrations into the equilibrium constant expression, \[K=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]}=\dfrac{(5.0 \times 10^{-2})^2}{(3.0 \times 10^{-3})^2(3.5 \times 10^{-3})}=7.9 \times 10^4 \nonumber \], To solve for \(K_p\), we use the relationship derived previously, \[K_p=7.9 \times 10^4 [(0.08206\; Latm/molK)(800 K)]^{1}\nonumber \], Hydrogen gas and iodine react to form hydrogen iodide via the reaction, \[H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\nonumber \], A mixture of \(H_2\) and \(I_2\) was maintained at 740 K until the system reached equilibrium. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large (\(K \geq 10^3\)). To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation \(\ref{Eq1}\)), for which K = 2.6 at 25C. I get that the equilibrium constant changes with temperature. B Substituting these values into the equation for the equilibrium constant, \[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78x)(0.21x)}=2.0 \times 10^{31}\nonumber \]. That's a good question! \(2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \), \(N_2O_{ (g)} + \dfrac{1}{2} O_{2(g)} \rightleftharpoons 2NO_{(g)} \), \(Cu_{(s)} + 2Ag^+_{(aq)} \rightleftharpoons Cu^{+2}_{(aq)} + 2Ag_{(s)} \), \(CaCO_{3 (g)} \rightleftharpoons CaCO_{(s)} + CO_{2 (g)} \), \(2NaHCO_{3 (s)} \rightleftharpoons Na_2CO_{3 (s)} + CO_{2 (g)} + H_2O_{ (g) }\). We can verify our results by substituting them into the original equilibrium equation: \[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{16})^2}{(0.78)(0.21)}=2.0 \times 10^{31}\nonumber \]. Calculate the equilibrium concentrations. Concentrations & Kc(opens in new window). We obtain the final concentrations by substituting this \(x\) value into the expressions for the final concentrations of n-butane and isobutane listed in the table: \[[\text{n-butane}]_f = (1.00 x) M = (1.00 0.72) M = 0.28\; M \nonumber \], \[[\text{isobutane}]_f = (0.00 + x) M = (0.00 + 0.72) M = 0.72\; M \nonumber \]. It's important to emphasize that chemical equilibria are dynamic; a reaction at . Keyword- concentration. Calculating Equilibrium Concentrations | Steps to Calculate | BYJU'S Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances. (Remember that equilibrium constants are unitless.). Direct link to Emily Outen's post when setting up an ICE ch, Posted 7 years ago. The equilibrium constant K for a system at equilibrium expresses a particular ratio of equilibrium constantBlank 1Blank 1 constant , Incorrect Unavailable of products and reactants at a particular temperatureBlank 2Blank 2 temperature , Correct Unavailable. Conversely, removal of some of the reactants or products will result in the reaction moving in the direction that forms more of what was removed. The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. We could solve this equation with the quadratic formula, but it is far easier to solve for \(x\) by recognizing that the left side of the equation is a perfect square; that is, \[\dfrac{x^2}{(0.0150x)^2}=\left(\dfrac{x}{0.0150x}\right)^2=0.106\nonumber \]. If this assumption is correct, then to two significant figures, \((0.78 x) = 0.78\) and \((0.21 x) = 0.21\). How is the Reaction Constant (Q) affected by change in temperature, volume and pressure ? Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium. The reaction quotient is calculated the same way as is \(K\), but is not necessarily equal to \(K\). In other words, chemical equilibrium or equilibrium concentration is a state when the rate of forward reaction in a chemical reaction becomes equal to the rate of backward reaction. 10.3: The Equilibrium Constant - Chemistry LibreTexts In the section "Visualizing Q," the initial values of Q depend on whether initially the reaction is all products, or all reactants. Experts are tested by Chegg as specialists in their subject area. At equilibrium, both the concentration of dinitrogen tetroxide and nitrogen dioxide are not changing with time. Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Very important to know that with equilibrium calculations we leave out any solids or liquids and keep gases. open bracket, start text, N, O, end text, close bracket, squared, equals, K, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, space, space, space, space, space, space, space, start text, T, a, k, e, space, t, h, e, space, s, q, u, a, r, e, space, r, o, o, t, space, o, f, space, b, o, t, h, space, s, i, d, e, s, space, t, o, space, s, o, l, v, e, space, f, o, r, space, open bracket, N, O, close bracket, point, end text, open bracket, start text, N, O, end text, close bracket, equals, square root of, K, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, end square root, 5, point, 8, times, 10, start superscript, minus, 12, end superscript, start text, M, end text, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket. The equation for the decomposition of \(NOCl\) to \(NO\) and \(Cl_2\) is as follows: \[2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber \], Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium. Check your answers by substituting these values into the equilibrium equation. Thus K at 800C is \(2.5 \times 10^{-3}\). Direct link to Sam Woon's post The equilibrium constant , Definition of reaction quotient Q, and how it is used to predict the direction of reaction, start text, a, A, end text, plus, start text, b, B, end text, \rightleftharpoons, start text, c, C, end text, plus, start text, d, D, end text, Q, equals, start fraction, open bracket, start text, C, end text, close bracket, start superscript, c, end superscript, open bracket, start text, D, end text, close bracket, start superscript, d, end superscript, divided by, open bracket, start text, A, end text, close bracket, start superscript, a, end superscript, open bracket, start text, B, end text, close bracket, start superscript, b, end superscript, end fraction, open bracket, start text, C, end text, close bracket, equals, open bracket, start text, D, end text, close bracket, equals, 0, open bracket, start text, A, end text, close bracket, equals, open bracket, start text, B, end text, close bracket, equals, 0, 10, start superscript, minus, 3, end superscript, start text, C, O, end text, left parenthesis, g, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, g, right parenthesis, \rightleftharpoons, start text, C, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, open bracket, start text, C, O, end text, left parenthesis, g, right parenthesis, close bracket, equals, open bracket, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, g, right parenthesis, close bracket, equals, 1, point, 0, M, open bracket, start text, C, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, close bracket, equals, open bracket, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, close bracket, equals, 15, M, Q, equals, start fraction, open bracket, start text, C, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, close bracket, open bracket, start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, close bracket, divided by, open bracket, start text, C, O, end text, left parenthesis, g, right parenthesis, close bracket, open bracket, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, g, right parenthesis, close bracket, end fraction, equals, start fraction, left parenthesis, 15, M, right parenthesis, left parenthesis, 15, M, right parenthesis, divided by, left parenthesis, 1, point, 0, M, right parenthesis, left parenthesis, 1, point, 0, M, right parenthesis, end fraction, equals, 225. Explanation: At equilibrium the reaction remains constant The rate of forward reaction equals rate if backward reaction Concentration of products and reactants remains same Advertisement ejkraljic21 Answer: The rate of the forward reaction equals the rate of the reverse reaction. Given: balanced equilibrium equation and values of \(K_p\), \(P_{O_2}\), and \(P_{N_2}\). Chapter 15 achieve Flashcards | Quizlet If the K value given is extremely small (something time ten to the negative exponent), you can elimintate the minus x in that concentration, because that change is so small it does not matter. Equilibrium constant are actually defined using activities, not concentrations. \([C_2H_6]_f = (0.155 x)\; M = 0.155 \; M\), \([C_2H_4]_f = x\; M = 3.6 \times 10^{19} M \), \([H_2]_f = (0.045 + x) \;M = 0.045 \;M\). Write the equilibrium equation. and products. C The small \(x\) value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table: We can verify our calculations by substituting the final concentrations into the equilibrium constant expression: \[K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{19})}=9.6 \times 10^{18}\nonumber \]. Moreover, we are told that at equilibrium the system contains 0.056 mol of \(Cl_2\) in a 2.00 L container, so \([Cl_2]_f = 0.056 \;mol/2.00 \;L = 0.028\; M\). There are three possible scenarios to consider: In this case, the ratio of products to reactants is less than that for the system at equilibrium. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. A \([CO_2]_i = 0.632\; M\) and \([H_2]_i = 0.570\; M\). Select all of the true statements regarding chemical equilibrium: 1) The concentrations of reactants and products are equal. 4) The rates of the forward and reverse reactions are equal. If a mixture of 0.257 M \(H_2\) and 0.392 M \(Cl_2\) is allowed to equilibrate at 47C, what is the equilibrium composition of the mixture? Thus \([NOCl]_i = 1.00\; mol/2.00\; L = 0.500\; M\). , Posted 7 years ago. with \(K = 9.6 \times 10^{18}\) at 25C. We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of \(x\). \[\ce{n-butane_{(g)} \rightleftharpoons isobutane_{(g)}} \label{Eq1} \]. Direct link to Zenu Destroyer of Worlds (AK)'s post if the reaction will shif, Posted 7 years ago. Because 1 mol of \(CO\) is produced for every 1 mol of \(H_2O\), the change in the concentration of \(CO\) is the same as the change in the concentration of H2O, so [CO] = +x.
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