Chemical Equation and Stoichiometry

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1 31 Introduction to Chemical Engineering Calculations Lecture 4.

2 4 What can we learn from a chemical equation? C7H O2 <======> 7 CO2 + 8 H2O 1. What information can we get from this equation? 2. What is the first thing we need to check when using a chemical equation? 3. What do you call the number that precedes each chemical formula? 4. How do we interpret those numbers? 2

3 What can we learn from a chemical equation? 1. A chemical equation provides both qualitative and quantitative information 2. Before using a chemical equation, make sure that it is balanced. 3. The number that precedes a compound is known as the stoichiometric coefficient. 4. The stoichiometric coefficient may be interpreted as number of moles or molecules. 3

4 The Stoichiometric Coefficient The stoichiometric coefficients tell us the mole (not mass) relationships (stoichiometric ratios) of the compounds participating in the reaction. For the chemical equation shown previously, 1 gmol of C 7 H 16 reacts with 11 gmol of O 2 to yield 7 gmol of CO 2 and 8 gmol of H 2 O 4

5 4 The Stoichiometric Coefficient Is it possible to establish the mass relationships from the stoichiometric coefficients? Consider: If 10 kg of C7H6 react completely with the stoichiometric quantity, how many kg of CO2 will be produced? 1kmol C7 H16 7 kmolco 2 44 kg CO 2 mco2 10 kg C7 H16 1kmolC H 1kmol CO 100 kg C H mco kg CO 2 5

6 Example Stoichiometry A limestone analysis shows the following composition: CaCO % MgCO % Inerts 1.70% a. How many pounds of calcium oxide can be made from 5 tons of this limestone? b. How many pounds of CO 2 can be recovered per pound of limestone? c. How many pounds of limestone are needed to make 1 ton of lime (mixture of CaO, MgO, and inerts)? 6

7 4 Example Stoichiometry Chemical Equations: CaCO3 <======> CaO + CO2 MgCO3 <======> MgO + CO2 Molecular Weights: CaCO3 MgCO3 CaO MgO CO

8 4 Example Stoichiometry Basis: 100 lbm of limestone Component i Mass (lbm) MW (lbm/lbmol) Mole (lbmol) CaCO MgCO Inerts Total

9 Example Stoichiometry Calculate the amount of products formed: 1lbmol CaO lbm CaO lbmol CaCO lbm CaO 3 1lbmol CaCO3 1lbmol CaO lbmol MgCO 1lbmol MgO 40.32lbm MgO 2.59 lbm MgO 3 1lbmol MgCO3 1lbmol MgO lbmol CaCO 3 1lbmol CO2 1lbmol CaCO lbm CO 1lbmol CO lbm CO lbmol MgCO 1lbmol CO 44.0 lbm CO lbm CO2 1lbmol MgCO3 1lbmol CO2 9

10 4 Example Stoichiometry Summary of Products Lime Source CO2 (lbm) Component Mass (lbm) CaCO3 CaO MgCO3 MgO Inerts Inerts Total 10

11 Example Stoichiometry Solution for (a): 2000lbm lime mcao 5tonsstone 5200lbm CaO 1ton 100lbmstone Solution for (b): 43.65CO2 lbm CO Ratio lbmstone lbmstone Solution for (c): 2000lbm 100lbmstone mstone 1ton lime 3550lbmstone 1ton 56.33lbm lime 11

12 Limiting and Excess Reactants Limiting Reactant The reactant that is present in the smallest stoichiometric amount. It is the compound that will be consumed first if the reaction proceeds to completion. Excess Reactant The chemical species whose amount supplied is higher than the stoichiometric requirement. The percentage excess is then computed as: actual amount present - theoretical amount needed % Excess = 100 theoretical amount needed 12

13 4 Limiting and Excess Reactants Consider a balanced chemical reaction: aa + bb ======> cc + dd Suppose x moles of A and y moles of B are present and they react according to the above reaction, x a if < then reactant A is limiting b y x a if > then reactant B is limiting b y 13

14 Conversion and Degree of Completion Conversion The fraction (or percentage) of the reactants that actually reacted during the reaction. amount of reactant converted % Conversion = 100 amount of reactant supplied Degree of Completion The fraction (or percentage) of the limiting reactant converted into products. amount of lim. reactant converted Degreeof Completion = 100 amount of lim. reactant supplied 14

15 Yield and Selectivity Yield The actual amount of products formed relative to what is actually expected from the reaction. actual amount of product formed %Yield = 100 theoretical amount of product expected Selectivity The ratio of the moles of the desired product produced to the moles of undesired product (by-product). moles of desired product formed Selectivity = moles of undesired product formed 15

16 4 Yield and Selectivity For example, methanol (CH3OH) can be converted into ethylene (C2H4) and propylene (C3H6) by the reaction: 2 CH3OH <======> C2H4 + 2 H2O 3 CH3OH <======> C3H6 + 3 H2O If the desired product is ethylene, then the selectivity is computed as: moles of ethylene formed Selectivity = moles of propylene formed 16

17 4 Example 4.2 Antimony (Sb) is obtained by heating pulverized stibnite (Sb2S3) with scrap iron and drawing off the molten antimony from the bottom of the reaction vessel 2 Sb2S3 + 3Fe <======> 2Sb + 3FeS Suppose that kg of stibnite and kg of iron turnings are heated together to give kg of Sb metal. Determine: a. b. c. d. e. The The The The The limiting reactant percentage of the excess reactant degree of completion (fraction) percent conversion of stibnite mass yield relative to stibnite supplied 17

18 4 Example 4.2 The molecular weights needed to solve the problem and the gmol forming the basis are: Component kg Mol. Wt. gmol Sb2S Fe Sb FeS

19 Example 4.2 Solution for (a): From chemical equation: Based on the actual supply: mol Sb2S mol Fe 3 stoichiometric mol Sb2S mol Fe 4.48 actual Since actual ration > stoichiometric ratio, hence, Fe is the limiting reactant and Sb 2 S 3 is the excess reactant. 19

20 Example 4.2 Solution for (b): The Sb2S3 required to react with the limiting reactant is 1gmol Sb S 4.48gmol Fe 3gmol Fe gmol Sb S 2 3 The percentage of excess reactant is gmol %excess Sb2S 3 = % 1.49gmol 20

21 Example 4.2 Solution for (c): Determine how much Fe actually does react: 3gmol Fe 1.64 gmol Sb 2.46gmol Fe 2gmol Sb The fractional degree of completion is 2.46gmol Fe fractional deg. of completion = gmol Fe 21

22 Example 4.2 Solution for (d): Determine how much Sb 2 S 3 actually does react: 1gmol Sb S 1.64gmol Sb 2gmol Sb gmol Sb S 2 3 The percentage conversion is 0.82gmol Sb S 2 3 %conversion Sb2S 3 = % 1.77gmol Sb2S 3 22

23 Example 4.2 Solution for (e): As required, the yield is expressed in terms of the mass of product (Sb) formed per mass of stibnite fed kg Sb kg Sb yield kg Sb S kg Sb S

24 Extent of Reaction () The extent of the reaction () is an extensive quantity describing the progress of a chemical reaction. Depends on the degree of completion of the reaction N i = N i0 v i where N i N i0 v i = molar amount remaining of species I = initial amount in the feed = stochiomettric coefficient (+ for product, for reactants) = extent of reaction 24

25 4 Equilibrium Constant, K Consider a reversible chemical reaction, aa + bb <======> cc + dd At equilibrium, There will be no net change in the amount of each chemical species. Hence, their individual concentrations are already constant. The equilibrium constant can be defined by, c k= d C D a b A B PCc PDd y cc y dd = a b= a b PA PB y AyB 25

26 4 Example 4.3 The gas-phase reaction between methanol and acetic acid to form methyl acetate and water takes place in a batch reactor and proceeds to equilibrium. CH3OH + CH3COOH <====> CH3COOCH3 + H2O (A) (B) (C) (D) When the reaction mixture comes to equilibrium, the mole fractions (y) of the four reactive species satisfy the relation y Cy D 4.87 y AyB 26

27 Example 4.3 a. If the feed to the reactor contains equimolar quantities of methanol and acetic acid and no other species, calculate the fractional conversion at equilibrium. b. It is desired to produce 70 mol of methyl acetate starting with 80 mol acetic acid. If the reaction proceeds to equilibrium, how much methanol must be fed? Assume no products are present initially. c. What is the composition of the final mixture at equilibrium in terms of the mole fractions? 27

28 Example 4.3 Solution to (a): Let n A, n B, n C, and n D be the respective molar quantities of A, B, C, and D at equilibrium. The total moles (n T ) is n T = n A + n B + n C + n D Expressing the equilibrium amounts in terms of initial amounts using the extent of reaction. n A = n A0 (1) n B = n B0 (1) n C = n C0 + (1) n D = n D0 + (1) 28

29 Example 4.3 The mole fractions at equilibrium are y A = n A /n T = (n A0 )/n T y B = n B /n T = (n B0 )/n T y C = n C /n T = (n C0 + )/n T y D = n D /n T = (n D0 + )/n T And the equilibrium constant can be written as nc0 n D0 n n A0 B

30 Example 4.3 For a basis of 1 gmol of A (initial amount) n A0 = n B0 = 1 gmol n C0 = n D0 = 0 Using these values in the equilibrium constant relation: Rearranging the equation: = 0 30

31 Example 4.3 Solving for : 1 = 1.83 gmol and 2 = gmol The last value is chosen since the first one results to a negative conversion. Solving for fractional conversion: X A amount of A reacted N N initial amount of A N N A A gmol XA gmol A0 A0 31

32 4 Example 4.3 Solution to (b): The initial amounts of the components are na0 = to be calculated and nb0 = 80 gmol nc0 = nd0 = 0 And the amounts at equilibrium are (1) na = na0 = na0 (2) nb = nb0 = (80 gmol) (3) nc = nc0 + = (0) + = 70 gmol (4) nd = nd0 + = (0) + 32

33 4 Example 4.3 From (3), = 70 gmol Updating the set of equations (1) na = na0 70 (2) nb = = 10 gmol (3) nc = = 70 gmol (4) nd = = 70 gmol Writing the equilibrium constant relation: nc n D 70gmol 70gmol 4.87 n A nb n A0 70gmol 10gmol 33

34 4 Example 4.3 Solving the last equation for na0: na0 = gmol of methanol Solution for (c): Component Moles at equilibrium Mole Fraction A na = gmol ya = B nb = 10 gmol yb = C nc = 70 gmol yc = D nd = 70 gmol yd = Total nt = gmol

THE CHEMICAL REACTION EQUATION AND STOICHIOMETRY

THE CHEMICAL REACTION EQUATION AND STOICHIOMETRY 9.1 Stoichiometry Stoichiometry provides a quantitative means of relating the amount of products produced by chemical reaction(s) to the amount of reactants. You should take the following steps in solving