Molarity (Concentration of Solutions) = M M= Moles of Solute = Liters of Solution mol L solute = material dissolved into the solvent In sea water, water is the solvent, and NaCl, MgCl2, etc are the solutes. In brass, copper is the solvent (90%), and zinc is the solute (10%). Preparing a Solution - I Example problem: A solution of sodium phosphate is prepared by dissolving 3.95 g of sodium phosphate in water and diluting it to 300.0 mL. What is the molarity, M, of the salt and each of the ions? Strategy (1)Write chemical equation showing process. (2)Calculate moles of each species. (3)Divide # moles by # L water to obtain molarity. (1)Na3PO4 (s) H2O(solvent) → 3 Na+ (aq) + PO4-3 (aq) Dilution of Solutions Dilute 25.00 mL of 0.0400 M KMnO4 to a final volume of 500. mL. What is the resulting molarity (M) of the diluted solution? Another Strategy for calculating final concentration: The number of moles of solute is the same before and after dilution V1 x M1 = moles solute = V2 x M2 V1M1 = V2 M 2 V1M1 M2 = V2 25.00mL × 0.0400M = 0.00200 M M2 = 500.mL TYPES of CHEMICAL REACTIONS in LIQUID SOLUTIONS: • Precipitation reactions • Acid – Base reactions • Reduction – Oxidation (REDOX) reactions Table 4.1 Simple Rules for Solubility of Salts in Water 1. Most nitrate (NO3-) salts are soluble. 2. Most salts of Na+, K+, and NH4+ are soluble. 3. Most chloride salts are soluble. Notable exceptions are AgCl, PbCl2, and Hg2Cl2. 4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, and CaSO4. 5. Most hydroxide salts are only slightly soluble. The important soluble hydroxides are NaOH, KOH, and Ca(OH)2 (marginally soluble). 6. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-) salts are only slightly soluble. Predicting if a precipitate forms, and which? Pb(NO3)2(aq) + NaCl(aq) → Pb+2(aq) + 2 NO3- (aq) + Na+ (aq) + Cl-(aq) RULE: If any of the possible new species formed by combining anions with cations is insoluble, then that precipitate will form. USE TABLE 4.1 (I’ll give you this Table for the tests, but not on quiz.) In this case, PbCl2 is insoluble, and a precipitate forms. Note: Like the demo with PbI2(s) above. Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Molecular Equation Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) +2 NaNO3 (aq) Complete Ionic Equation Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3-(aq) Net Ionic Equation Ca2+(aq) + SO4-2(aq) Spectator Ions are Na+ and NO3- CaSO4 (s) Quantitative Precipitation Problems: Calculate the mass of solid sodium iodide that must be added to 2.50 L of a 0.125 M lead nitrate solution to precipitate all of the lead as PbI2 (s)! The chemical equation for the reaction is: Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq) The moles of sodium iodide needed to precipitate PbI2 is twice the lead ions. The number of moles of sodium iodide needed is: 2+ 0.125 Mol Pb 2 mol I 2.50 L x x 1.0 L soln. 1 mol Pb2+ The mass of sodium iodide is: 0.625 mol I- x 1 mol NaI x 1 mol I- = 0.625 mol I- 149.9 g NaI = 93.68 g NaI 1 mol NaI Precipitation problem: When aqueous silver nitrate and sodium chromate solutions are mixed, solid silver chromate forms in a solution of sodium nitrate. If 257.8 mL of a 0.0468 M solution of silver nitrate is added to 156.00 mL of a 0.0950 M solution of sodium chromate, what mass of silver chromate (M = 331.8 g/mol) will be formed? This is a limiting-reactant problem because the amounts of two reactants are given. Strategy: (1) Write the balanced equation. (2) Calculate the number of moles of each reactant. (3) Determine the limiting reactant. (4) Calculate the moles of product. (5) Convert moles of product to mass of the product using molar mass. Selected Acids and Bases Acids Bases Strong: H+(aq) + A-(aq) Hydrochloric, HCl Hydrobromic, HBr Hydroiodoic, HI Nitric acid, HNO3 Sulfuric acid, H2SO4 Perchloric acid, HClO4 Strong: M+(aq) + OH-(aq) Sodium hydroxide, NaOH Potassium hydroxide, KOH Calcium hydroxide, Ca(OH)2 Strontium hydroxide, Sr(OH)2 Barium hydroxide, Ba(OH)2 Weak Hydrofluoric, HF Phosphoric acid, H3PO4 Acetic acid, CH3COOH (or HC2H3O2) Weak Ammonia, NH3 accepts proton from water to make NH4+(aq) and OH-(aq) Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq) Moles H2SO4 = 155 g H2SO4 x 1 mole H2SO4 = 1.58 moles H2SO4 98.09 g H2SO4 -2 1.58 mol SO 4 Molarity of SO4- 2 = = 0.687 Molar in SO4- 2 2.30 L solution Molarity of H+ = 2 x 0.687 M = 1.37 Molar in H+ (or H3O+) Writing Balanced Equations for Neutralization Reactions - I Problem: Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following Chemical reactions: a) Calcium Hydroxide(aq) and Hydroiodic acid(aq) b) Lithium Hydroxide(aq) and Nitric acid(aq) c) Barium Hydroxide(aq) and Sulfuric acid(aq) Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts. Solution: a) Ca(OH)2 (aq) + 2HI(aq) CaI2 (aq) + 2H2O(l) Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq) Ca2+(aq) + 2 I -(aq) + 2 H2O(l) 2 OH -(aq) + 2 H+(aq) 2 H2O(l) Like Example 4.10 (P 113) What volume of 0.468 M H2SO4 is needed to neutralize 215.00 ml of a 0.125 M LiOH solution? Calculate the number of moles of base: Vbase x Mbase = 0.21500 L x 0.125 M = 0.0268 mol LiOH From the balance equation find the moles of acid needed: 2 LiOH(aq) + H2SO4 (aq) 2 H2O(l) + Li2SO4 (aq) Since there are two protons per molecule, we will need half as much sulfuric acid as we have lithium hydroxide: or 0.0134 mol H2SO4 Volume of acid: Moles acid 0.0134 moles = 0.468 Mol = 0.0286 L H2SO4 Vacid = Macid L EP 91: Aluminum hydroxide reacts with hydrochloric acid According to the balanced equation Al(OH)3 (s) + 3 HCl (aq) → 3 H2O(l) + AlCl3 (aq) What volume of 1.50 M HCl(aq) is required to neutralize 10.0 g Al(OH)3(s)? Strategy: (1) Calculate moles of Al(OH)3(s). (2) Calculate moles of HCl needed using balanced equation (3) Calculate volume HCl from (2) and known molarity Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each of the Rx: a) Zn(s) + 2 HCl(aq) ZnCl2 (aq) + H2 (g) b) S8 (s) + 12 O2 (g) 8 SO3 (g) c) NiO(s) + CO(g) Ni(s) + CO2 (g) Plan: First we assign an oxidation number (O.N.) to each atom (or ion) based on the rules in Table 4.3. The reactant is the reducing agent if it contains an atom that is oxidized (O.N. increased in the reaction). The reactant is the oxidizing agent if it contains an atom that is reduced ( O.N. decreased). Solution: a) Assigning oxidation numbers: -1 +1 0 Zn(s) + 2 HCl(aq) -1 0 +2 ZnCl2 (aq) + H2 (g) HCl is the oxidizing agent, and Zn is the reducing agent! Problem: Calculate the mass of metallic Iron that must be added to 500.0 liters of a solution containing 0.00040M of Pt2+(aq) ions in solution to reclaim all Pt via: 2 Fe(s) + 3 Pt2+(aq) → 2 Fe3+(aq) + 3 Pt(s) Solution: V x M = # moles 500.0L x 0.00040 mol Pt2+/ L = 0.20 mol Pt2+ Fe(s) → Fe3+ + 3 ePt2+ + 2 e- → Pt(s) Need 2 moles of Iron for every 3 moles of Platinum 0.20 mol Pt2+ 2 mol Fe x = 0.133 mol Fe 2+ 3 mol Pt 0.133 mol Fe x 55.85 g Fe = 7.4 g Fe mol Fe Balancing REDOX Equations: The oxidation number method Step 1) Assign oxidation numbers to all elements in the equation. Step 2) From the changes in oxidation numbers, identify the oxidized and reduced species. Step 3) Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. Draw tie-lines between these atoms to show electron changes. Step 4) Choose coefficients for these species to make the electrons lost equal the electrons gained (or total increase in ON = total decrease in ON) Step 5) Complete the balancing by inspection. REDOX Balancing Using Ox. No. Method - II +2 -1e- +3 Fe+2(aq) + MnO4-(aq) + H+(aq) Fe+3(aq) + Mn+2(aq) + H2O(aq) +5 e+2 +7 Balance charge: Multiply Fe+2 & Fe+3 by five to correct for the electrons gained by the Manganese: 5 Fe+2(aq) + MnO4-(aq) + H+(aq) 5 Fe+3(aq) + Mn+2(aq) + H2O(aq) Balance O: Need 4 H2O on right to cancel O from the MnO4-. Balance H: Need 8 H+ on the left to balance these water Hs. 5 Fe+2(aq) + MnO4-(aq) +8 H+(aq) 5 Fe+3(aq) + Mn+2(aq) +4 H2O(aq) BALANCED! Note: Add 8 water molecules to both sides if you prefer to express the 8 H+ ions instead as H3O+ ions. Balancing redox eqn using half-cell method in acidic solutions Cu(s) +HNO3(aq) → Cu2+(aq) + NO(g) Identify half-reactions, one is ox, other is red Cu(s) → Cu2+(aq) 0 → +2 HNO3(aq) → NO(g) +5 → +2 Balance all atoms that are neither H nor O OK as is Balance O by adding H2O to side deficient in O Cu(s) → Cu2+(aq) HNO3(aq) → NO(g) + 2H2O Balance H by adding H+ to side deficient in H Cu(s) → Cu2+(aq) 3H+ +HNO3(aq) → NO(g) + 2H2O Balance charge by adding e- to side that has + charge Cu(s) → Cu2+(aq) + 2e3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O Multiply each eq by factors so electrons cancel out 3x(Cu(s) → Cu2+(aq) + 2e-) 2x(3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O) Add equations 3Cu(s)+ 6H+ + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O Balancing redox equations in basic solutions: First balance in acidic solution, then add OH- to cancel H+ The following redox equation is balanced in acidic solution. Balance it for a basic solution. Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3- (aq) +3H2O(l) Add OH- equal to number of H+ on both sides 6OH-(aq) +Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3-+3H2O(l) +6OH-(aq) Combine OH- and H+ to form water to max extent possible 6H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3-+3H2O(l) +6OH- (aq) Cancel H2O on both sides to max extent possible 3H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3- +6OH- (aq) Redox Titration- Calculation outline - I Volume (L) of KMnO4 Solution a) M (mol/L) Moles of KMnO4 b) Molar ratio Moles of CaC2O4 c) Chemical Formula Moles of Ca+2 Problem: Calcium Oxalate was precipitated from blood by the addition of Sodium Oxalate so that calcium ion could be determined. In the blood sample. The sulfuric acid solution that the precipitate was dissolved in required 2.05 ml of 4.88 x 10-4 M KMnO4 to reach the endpoint. a) calculate the amount (mol) of Ca+2. b) calculate the Ca+2 ion conc. Plan: a) Calculate the molarity of Ca+2 in the H2SO4 solution. b) Convert the Ca+2 concentration into units of mg Ca+2/ 100 ml blood. Ch. 5: Pressure = force per unit area Density of Mercury = 13.6 g/cm3 760 mm column of 1 cm2 area weighs 76 cm x 1 cm2 x 13.6 g/cm3 = 1030 g = 1.03 kg = 2.28 lbs P = force / area = 2.28 pounds / cm2 = 14.7 pounds / in2 = 1.00 atm. Force = weight = volume x density = area x height x density P = Force / area = height x density IDEAL GAS LAW • The ideal gas equation combines both Boyles Law and Charles Law into one easy-to-remember law: PV=nRT • • • • n = number of moles of gas in volume V R = Ideal gas constant R = 0.08206 L atm / (mol K) = 0.08206 L atm mol-1 K-1 Later R = 8.314 J / (mol K) = 8.314 J mol-1 K-1 An ideal gas is one for which both the volume of molecules and forces between the molecules are so small that they have insignificant effect on its P-V-T behavior. Independent of substance, in the limit that n/V →0, all gases behave ideally. Usually true below 2 atm. Variations on Ideal Gas Equation • • During chemical and physical processes, any of the four variables in the ideal gas equation may be fixed. Thus, PV=nRT can be rearranged for the fixed variables: – for a fixed amount at constant temperature • P V = nRT = constant Boyle’s Law – for a fixed amount at constant pressure • V / T = nR / P = constant Charles’ Law – for a fixed pressure and temperature • V = n (RT/P) or V/n = constant Avogadro’s Law – for a fixed amount at constant volume • P / T = nR / V = constant Amonton’s Law JUST REMEMBER: PV=nRT and rearrange as needed Standard Temperature and Pressure (STP) A set of Standard conditions have been chosen to make it easier to quantify gas amounts (i,.e., “liters at STP”). Standard Temperature = 00 C = 273.15 K Standard Pressure = 1 atmosphere = 760 mm Mercury At these “standard” conditions, if you have 1.0 mole of any ideal gas, it will occupy a “standard molar volume”: Standard Molar Volume = 22.414 Liters = 22.4 L Gas Law: Solving for Pressure Problem: Calculate the pressure in a container whose Volume is 87.5 L and it is filled with 5.038kg of Xenon at a temperature of 18.8 oC. Plan: Convert all information into the units required, and substitute into the Ideal Gas equation ( PV=nRT ). Solution: 5038 g Xe = 38.37014471 mol Xe nXe = 131.3 g Xe / mol T = 18.8 oC + 273.15 K = 291.95 K PV = nRT so P = nRT V P = (38.37 mol )(0.0821 L atm)(291.95 K) = 10.5108 atm = 10.5 atm 87.5 L (mol K) Applying the Gas law to T changes Problem: A copper tank is filled with compressed gas to a pressure of 4.28 atm at a temperature of 0.185 oF. What will be the pressure if the temperature is raised to 95.6 oC? Plan: The volume of the tank is not changed, and we only have to deal with the temperature change, so convert to SI units, and calculate the pressure ratio from the T ratio. Solution: P1 = P2 = nR/V T1 = (0.185 oF - 32.0 oF)x 5/9 = -17.68 oC T1 T2 T = -17.68 oC + 273.15 K = 255.47 K 1 T2 = 95.6 oC + 273.15 K = 368.8 K P2 = 4.28 atm x 368.8 K = 6.18 atm 255.47 K P2 = P1 x T2 = ? T1 Change of Conditions :Problem -I • A gas sample in the laboratory has a volume of 45.9 L at 25 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by pumping (compressing) the gas to a new volume of 3.10 ml what is the pressure? • • • • • P1= 743 mm Hg x1 atm/ 760 mm Hg=0.978 atm P2 = ? V1 = 45.9 L V2 = 3.10 ml = 0.00310 L T1 = 25 oC + 273 = 298 K T2 = 155 oC + 273 = 428 K Example Problem: Molar Mass of a Gas from its weight and P,V,T Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 ml flask. If the sample had a mass of 0.118 g at a pressure of 550.0 Torr, what is the molecular weight of the gas? Plan: Use the Ideal gas law to calculate n, then calculate the molar mass. Solution: P = 550.0 Torr x 1mm Hg x 1.00 atm = 0.724 atm 1 Torr 760 mm Hg V = 250.0 ml x 1.00 L = 0.250 L 1000 ml n =P V RT T = 25.0 oC + 273.15 K = 298.2 K n = (0.724 atm)(0.250 L) = 0.007393 mol (0.0821 L atm/mol K) (298.2 K) M = mass / n = 0.118 g / 0.007393 mol = 16.0 g/mol Ideal Gas Mixtures • Ideal gas equation applies to the mixture as a whole and to each gas species (i) individually: PtotalV = ntotalRT Ptotal is what is measured by pressure guages. Σ ni = n1+n2+n3…= ntotal i PiV = niRT for all species i. Pi much harder to measure. Note: iΣ Pi = P1+P2+P3…= Ptotal Chemical Equation Calc - III Mass Atoms (Molecules) Avogadro’s Number 6.02 x 1023 Reactants Molarity moles / liter Solutions Molecules Moles Molecular g/mol Weight Products PiV = niRT Gases Mole fraction of species i: xi = ni/ntotal • PiV = niRT for all i, and Σ Pi = P1+P2+P3…= Ptotal • Σ ni = n1+n2+n3…= ntotal • Σ xi = x1+x2+x3…= 1.00000 • Pi = niRT/V = (ni/ntotal)ntotalRT/V = xi . Ptotal Gas Law Stoichiometry - I - NH3 + HCl Problem: A slide separating two containers is removed, and the gases are allowed to mix and react. The first container with a volume of 2.79 L contains Ammonia gas at a pressure of 0.776 atm and a temperature of 18.7 oC. The second with a volume of 1.16 L contains HCl gas at a pressure of 0.932 atm and a temperature of 18.7 oC. What mass of solid ammonium chloride will be formed, and what will be remaining in the container, and what is the pressure? Plan: This is a limiting reactant problem, so we must calculate the moles of each reactant using the gas law to determine the limiting reagent. Then we can calculate the mass of product, and determine what is left in the combined volume of the container, and the conditions. NH3 HCl Solution: Equation: NH3 (g) + HCl(g) TNH3 = 18.7 oC + 273.15 = 291.9 K NH4Cl(s) Velocity and Energy • Kinetic Energy = KE = ½ mu2 for one molecule • Average Kinetic Energy of a mole of gas (KEavg) = 3/2 RT independent of gas identity • Average Kinetic Energy of one molecule (KEavg) = 3/2 RT/NA independent of mass, identity = ½ mu2 (mean value of u2) u2 = 3RT/(mNA) = 3RT/M √(u2) = √(3RT/M) “Root mean square velocity” Must use R in J/mol K, and remember that 1 J = 1 kg m2/s2 Molecular picture of HEAT! Diffusion Rates: proportional to average velocity or to 1/√M • Rate1/Rate2 = u1/u2 = (M2/M1)1/2 • HCl = 36.46 g/mol NH3 = 17.03 g/mol • RateNH3 = RateHCl x ( 36.46 / 17.03 )1/ 2 • RateNH3 = RateHCl x 1.463 The Equilibrium Constant - Definition Consider the generalized chemical reaction: aA+bB = cC+dD A, B, C and D represent chemical species and a, b, c, and d are their stoichiometric coefficients in the balanced chemical equation. At equilibrium, C ] [ D] [ K= a b [ A] [ B ] c d Note: The “units” for K are concentration units raised to some power = c+d–(a+b) The square brackets indicate the concentrations of the species in equilibrium and K is a constant called the equilibrium constant. K depends only on T , and not on concentrations. 2 [ ][ ] CO SO 2 2 CO2(g) + 2 SO2(g)K = 1 3 [CS2 ][O2 ] CS2(g) + 3 O2(g)R • The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original expression. 3 [ CS2 ][O 2 ] CS2(g) + 3 O2(g)K 2 = [CO 2 ][SO 2 ]2 CO2(g) + 2 SO2(g)R • If the original reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power n. 2 CS2(g) + 6 O2(g) = 2 CO2(g) + 4 SO2(g) CO 2 ] [SO2 ] [ K3 = 2 6 [CS2 ] [O2 ] 2 4 = ( K1 ) 2 1 = K1 Example: Calculation of the Equilibrium Constant from equilibrium amounts At 454 K, the following reaction takes place: 3 Al2Cl6(g) R 2 Al3Cl9(g) At this temperature, the equilibrium concentration of Al2Cl6(g) is 1.00 M and the equilibrium concentration of Al3Cl9(g) is 1.02 x 10-2 M. Compute the equilibrium constant at 454 K. Strategy: Substitute values into K [ Al 3Cl9 ] (1.02 × 10 M ) K= = 3 3 (1.00 M ) [Al2Cl6 ] 2 −2 2 −4 = 1.04 × 10 M −1 Determining Equilibrium Concentrations from K Example: Methane can be made by reacting carbon disulfide with hydrogen gas. K for this reaction is 27.8 (L/mol)2 at 900°C. CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g) At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol CS2, 1.10 mol of H2, and 0.45 mol of H2S. How much methane was formed? Strategy: (1)Calculate the equilibrium concentrations from the moles given and the volume of the container. (2)Use the value of K to solve for the concentration of methane. (3) Calculate the number of moles of methane from M and V. Equilibrium Expressions Involving Pressures For a reaction of the type aA+bB = cC+dD It is sometimes convenient to write the equilibrium expression in terms of partial pressures, e.g. PC ) ( PD ) ( KP = a b ( PA ) ( PB ) c d = K ( RT ) P indicates the partial pressures of the species in equilibrium and KP is a constant called the equilibrium constant in terms of partial pressures. KP depends only on T , and not on pressure. Δn Equilibrium involving pure solids or liquids: set their activity = 1 Example: NH4NO2(s) = N2(g) + 2 H2O(g) The equilibrium constant for this reaction would normally be expressed as: 2 N 2 ][ H 2 O ] [ K'= [ NH 4 NO2 ] However, a pure solid or liquid retains the same activity during the reaction. Thus we set the activity of NH 4 NO 2 (s) to one. K = [ N 2 ][ H 2 O ] 2 1.0000 and ( )( P ) K p = PN2 H2O 1.0000 2 The Reaction Quotient, Q Consider the reaction: a A(g ) + b B(g ) = c C( g ) + dm D( g ) The reaction quotient, Q is defined as C ]t [ D ]t [ Q= a b [ A]t [ B ]t c d where the subscripts t indicate momentary concentrations at some time t before (or after) equilibrium has established. Q has same form as K, but the concentrations are the actual rather than the equilibrium concentrations Example: Calculating Equilibrium Pressures and Concentrations from K and initial conditions Consider the equilibrium: CO(g) + H2O(g) = CO2(g) + H2(g) 0.250 mol CO and 0.250 mol H2O are placed in a 125 mL flask at 900 K. What is the composition of the equilibrium mixture if K = 1.56? The original reactant concentrations are: [CO]0 = [H2O]0 = 0.250 mol/ 0.125 L = 2.00 M Q = 0. Therefore, Q < K, so reactants are consumed and products made. CO(g) + H2O(g) = CO2(g) + H2(g) Conc. (M) CO(g) H2O(g) CO2(g) H2(g) Init. 2.00 2.00 0 0 Change -x -x +x +x Equil. 2.00 - x x x 2.00 - x
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