K eq. reverse. b) 4 HCl (g) + O 2(g) 2 H 2 O (g) + 2 Cl 2(g) c) NOCl (g) NO (g) + ½ Cl 2(g) d) Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq)

Similar documents
Page 2. PhysicsAndMathsTutor.com

0620 CHEMISTRY. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers.

0620 CHEMISTRY. 0620/23 Paper 2 (Core Theory), maximum raw mark 80

3a. The acid and base react to form a salt solution of ammonium propionate. CaCO s + 2 HC H O aq CO g + H O l + Ca ( aq ) + 2 C H O ( aq)

Gas Tests. Distinguishing Chemical Property acts as the fuel for combustion, allowing things to burn more vigorously in its presence

Copyright 2016 Dan Dill 1

Soil Composition. Air

Chemical Equations Part 1

Zeolite Clinoptilolite Powder

Selected Water Quality Topics Related to Larval Shrimp Culture

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education CO-ORDINATED SCIENCES

ELEMENTAL BIOAVAILIBILITY IN NUTRIENT SOLUTIONS IN RELATION TO PRECIPITATION REACTIONS

TRATION: ANALYSIS OF SODIUM HYDROXIDE

Cambridge International Examinations Cambridge International General Certificate of Secondary Education

Chemistry 130. Acid and Base equilibria

Chinese Zinc Sulfate Monohydrate testing. Dick Camp Kronos Micronutrients

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education

Chlorination Behaviors of Zinc Phases by Calcium Chloride in High Temperature Oxidizing-chloridizing Roasting

PHOTOSYNTHESIS (7.5A)

CH 3 CH 2 CH 2 CH 2 OH

People who have a zinc deficiency can take hydrated zinc sulfate (ZnSO 4.xH 2O) as a dietary supplement.

Copyright 2018 Dan Dill 1

WHAT ARE FERTILIZERS

Example: Ammonium Sulphate (also called Sulphate of Ammonia) is composed of the following:

European Pharmacopoeia solutions

ZeoMed and Zeofresh and EcoFresh Ranges of Products SAFETY DATA SHEET

TNPSC Chemistry Study Material Fertilizers

FEEDBACK TUTORIAL LETTER TEST 2 SECOND SEMESTER 2018 BASIC SCIENCE [BSC410S]

EXPERIMENT 13 QUALITATIVE ANALYSIS

Pharmaceutical Analytical Chemistry PHCM223 Lecture 12 Applications on different types of equilibria. Dr. Nesrine El Gohary 12 th lecture

Skills and competencies

Strength of Vinegar by Acid-Base Titration

Functional Derivatives of Carboxylic Acids

CARBOXYLIC ACIDS AND THEIR DERIVATIVES: NUCLEOPHILIC ADDITION-ELIMINATION AT THE ACYL CARBON

Ch 2 Molecules of life

Homework Chapter 08 Chemistry 51 Los Angeles Mission College

Plant Nutrients in Mineral Soils

1/3/2011. Chapter 17 Carboxylic Acids and Their Derivatives. Nucleophilic Addition- Elimination at the Acyl Carbon

5124 SCIENCE (PHYSICS AND CHEMISTRY)

SAC 17 Queen Mary University of London

AGRY 515: What do you know? In 10 minutes, fill out what you can. Educated guesses are strongly encouraged.

By Andrew & Erin Oxford, Bethel

EXPERIMENT 4 TITRATION OF AN UNKNOWN ACID

G/LITRE 5.0 g KOH g 0.5 g 0.05 g 0.01 g MgS047H20 NaCl CaCl2

Question Number. Answer Acceptable answers Mark. 1(a)(i) A calcium ion, Ca 2+ (1) Question Number. Answer Acceptable answers Mark (2)

CHAPTER4 ANSWERS. Multiple Choice Questions. Short Answer Questions. 1. (b) 2. (d) 3. (a) 4. (c) 5. (c) 6. (b) 7. (a) 8. (b)

Agronomy 365 Exam I Spring 2002

Essential plant nutrients. Classification of essential plant nutrients

Soils, Fertilizers and Plant Nutrition 1

TECHNICAL SPECIFICATION No. 5 PRILLED NPK COMPLEX CHEMICAL FERTILIZERS

Fundamentals of Organic Chemistry CHEM 109 For Students of Health Colleges Credit hrs.: (2+1)

IODOMETRIC TITRATION

Tissue culture Media preparation

1 Small molecules are used as the basic units in the synthesis of large food molecules. Which statement is correct? A

(Writing model for laboratory note book)

EXPERIMENT #5 Water of Hydration

XI CLASS BIOLOGY CHAPTER 12: MINERAL NUTRITION

Warm-Up. Distinguish between an element and a compound. Element. Compound

Soil Fertility and Nutrient Management. Hailin Zhang. Department of Plant and Soil Sciences

Soils and Soil Fertility Management

Fertilizers. Chapter INTRODUCTION

H y., are burned in 100 cm 3 of oxygen, which is an excess of oxygen.

Supplementary Figure 1. Distribution of the study lakes across Sweden.

3.12 Polymers. As with esters, the OH from the carboxylic acid is lost with the H from the alcohol.

Experiment 10 Acid-base Titrations: Part A Analysis of vinegar and Part B Analysis of a Carbonate/Bicarbonate mixture

Chapter-2 Mineral Nutrition

Phosphate removal from secondary effluent of wastewater treatment: characterization and potential re-use as fertilizer of recovered precipitates

Supplying Nutrients to Crops

Chapter 18. Carboxylic Acids and Their Derivatives. Nucleophilic Addition-Elimination at the Acyl Carbon

Importance of fertigation scheduling

Soil organic matter composition, decomposition, mineralization and immobilization

Maharashtra State Board Class IX Science and Technology Sample Paper 3

Classes at: - Topic: Redox & Volumetric Titration

The role of water in the maintenance of life

PRECIPITATON TITRATION

Identification and qualitative Analysis. of Renal Calculi

Biosolids Test Description Method Number Test Code Acid Hydrolyzable Phosphate SM P E-99 AHPHOS Acidity as CaCO3 SM-2310 B (4a)-97 ACID

Phosphorus recovery from human urine

Fertilization Programming

Terry Richmond s Fertilizer Package mentioned in the panel discussion March 14, 2013.

THIONYL IODIDE. Part II. Rate of Decomposition and Spectroscopic Studies BY M. R. ASWATHANARAYANA RAO. Introduction

EXPERIMENT 14 ANALYSIS OF OILS AND FATS:

Magnesium oxide: A Forgotten Specialty Chemical By Marcelo Fraga de Sousa* Buckman, Sumaré, SP - Brazil

Titration of Synthesized Aspirin A continuation of the aspirin synthesis lab

TRATION: ANALYSIS OF SODIUM HYDROXIDE

III. TOXICOKINETICS. Studies relevant to the toxicokinetics of inorganic chloramines are severely

10. CARBOXYLIC ACIDS AND THEIR DERIVATIVES 10.1 Nomenclature of Carboxylic Acids 10.2 Physical Properties of Carboxylic Acids 10.

Class 10 Science - Sample Paper Set I

Kirkcaldy High School - Chemistry Higher Assignment Pupil Guide. Antioxidants

4 Biological Molecules

Fertilizer. Fertilizers. (FAO press release, April 1990): Nutrient Depletion. Nutrient Depletion 4/16/2012

Limitations to Plant Analysis. John Peters & Carrie Laboski Department of Soil Science University of Wisconsin-Madison

Basic Consideration on EAF Dust Treatment Using Hydrometallurgical Processes

Plants, soil, and nutrients. Created in partnership with Alex Lindsey, Ph.D., The Ohio State University

Visit For All NCERT solutions, CBSE sample papers, Question papers, Notes for Class 6 to 12. Chapter-12 MINERAL NUTRITION

1. 09/07/16 Ch 1: Intro to Human A & P 1

Fluid and Electrolytes P A R T 4

Experiment: Iodometric Titration Analysis of Ascorbic Acid Chem251 modified 09/2018

Transcription:

Name: 1 Equilibrium Worksheet SOLUTIONS Complete the following questions on a separate piece of paper. 1. Write the uilibrium epression,, for each of the following reactions: a NO (g + O (g NO (g [ NO [ NO [ O b HCl (g + O (g H O (g + Cl (g O [ Cl Cl [ O c NOCl (g NO (g + ½ Cl (g 1 [ NO [ Cl [ NOCl d Fe + (aq + SCN - (aq FeSCN + (aq [ FeSCN [ Fe [ SCN e CaCl (s Ca + (aq + Cl - (aq [ Ca [ Cl f HC H O (aq + H O (l H O + - (aq + C H O (aq O [ CH O C H O g MgO (s + CO (g MgCO (s 1 [ CO h C (s + CO (g + Cl (g COCl (g [ COCl [ CO [ Cl i Ca (PO (s Ca + - (aq + PO (aq [ Ca [ PO j ZnO (s + CO (g Zn (s +CO (g [ CO [ CO. The uilibrium constant for the uilibrium CO (g + H O (g CO (g + H (g is 0 at 600. What is the value of the uilibrium constant for the reverse reaction at the same temperature? 1 1 reverse.110 0. Classif the following uilibria as heterogeneous or homogeneous, and write an uilibrium epression for each. a NH NO (s N (g + H O (g heterogeneous [ N O

Name: b H O (l H O (g heterogeneous H O [ ( g c SO (g + ½ O (g SO (g homogeneous [ SO 1/ [ SO [ O d S 8(s + 8 O (g 8 SO (g heterogeneous 8 [ SO 8 [ O. At the uilibrium point in the decomposition of phosphorus pentachloride to chlorine and phosphorus trichloride, the following concentrations are obtained: 0.010 mol/l PCl 5, 0.15 mol/l PCl and 0.7 mol/l Cl. Determine the for the reaction. PCl 5(g PCl (g + Cl (g C = 0.001 M 0.15 M 0.7 M [ PCl [ Cl [ PCl 5 (0.15(0.7 (0.010 5.55 Therefore the value for the reaction is 5.55. 5. The colourless gas dinitrogen tetroide decomposes to the brown coloured air pollutant nitrogen dioide and eists in uilibrium. A 0.15 mol sample of dinitrogen tetroide is introduced into a 1.00 L container and allowed to decompose at a given temperature. When uilibrium is reached, the concentration of the dinitrogen tetroide is 0.0750 mol/l. What is the value of for this reaction? N O (g NO (g MR 1 I 0.15 0 C - + E 0.15- =0.0750 =0.050 [ NO [ N O (0.10 (0.075 0.1 Therefore the value for the reaction is 0.1 6. Phosphorus pentachloride decomposes to phosphorus trichloride and chlorine gas. A 1.10 mol/l sample of PCl 5 was placed into a vessel. At uilibrium it was determined that the concentration was 0. mol/l. Calculate the for this reaction. PCl 5(g PCl (g + Cl (g [ PCl [ Cl MR 1 1 1 [ PCl5 I 1.10 0 0 (0.77(0.77 C - + + (0. E 1.10- =0. 1.80 = 0.77 Therefore the value for the reaction is 1.80

Name: 7. Gaseous dinitrogen tetroide is placed in a flask and allowed to decompose to nitrogen dioide and reach uilibrium at 100 C. At 100 C, the value of is 0.1. If the concentration of dinitrogen tetroide at uilibrium is 0.155 mol/l, what is the concentration of nitrogen dioide at uilibrium? N O (g NO (g 0.1 [ NO [ NO [ N O [ NO (0.155 0.181M 8. A 0.91 mol sample of dinitrogen tetroide is placed in a 1.00 L vessel and heated to 100 C. At uilibrium it is found that 0.7 % of the dinitrogen tetroide has decomposed to nitrogen dioide. Calculate the for this reaction. N O (g NO (g MR 1 I 0.91 0 C - + E 0.91- At uilibrium, 0.7% of N O is gone 0.00 = 0.7% of 0.91 = (0.07(0.91 = 0.191 [N O = 0.70 M [NO = 0.8 M Therefore the uilibrium constant is 0.00. 9. At 5 C, the uilibrium concentration of dinitrogen tetroide gas is 6.810 - mol/l and the total gas concentration is 1.10 - mol/l. Determine the for the decomposition of dinitrogen tetroide gas to nitrogen dioide gas at this temperature. N O (g NO (g [N O = 6.810 - M [gas = 1. 10 - M [NO = 1. 10 - M - 6.810 - M = 5.910 - M [ NO [ N O (5.910 (6.810 5.910 [ NO [ N O (0.8 (0.70 - Therefore the uilibrium constant is 5.9X10 -.

Name: 10. The following reaction has value of 85.0 at 60 C: SO (g + NO (g NO (g + SO (g If a miture of sulfur dioide and nitrogen dioide is prepared, each with an initial concentration of 0.100 mol/l, calculate the uilibrium concentrations of nitrogen dioide and nitrogen monoide at this temperature. SO (g + NO (g NO (g + SO (g = 85.0 MR 1 1 1 1 I 0.100 0.100 0 0 C - - + + E 0.100-0.100- [ NO[ SO [ SO [ NO [NO = 9.810 - M 85 9. 0.9 ( ( (0.100 (0.100 (0.100 0.090 10. [NO = 0.090 M Therefore the concentration of nitrogen dioide is 9.810 - M and nitrogen monoide is 0.090 M. 11. At 100 C the reaction below has an uilibrium constant,, value of.10-10. If 1.00 mol of phosgene, COCl, is placed in a 10.0 L flask, calculate the concentration of carbon monoide at uilibrium. COCl (g CO (g + Cl (g 0.100/ < 1000 appro will not work COCl (g CO (g + Cl (g =.10-10 MR 1 1 1 I 0.100 0 0 C - + + E 0.100- X 0.100/ > 1000 appro will work Therefore [CO =.710-6 M.10.10 [ CO[ Cl [ COCl 10 10.710 6 ( ( 0.100 0.100 1. Si moles of SO (g and four moles of O (g are introduced into a 1.00 L reaction vessel and allowed to react to form SO (g. At uilibrium, the vessel contains four moles of SO (g. Calculate for this reaction. SO (g + O (g SO (g [O =.00 M MR 1 [ SO I 6.00.00 0 [ SO [ O C - - + (.00 E 6.00-.00- =.00 At uilibrium [SO =.00 M (.00 (.00 Therefore =.00 M [SO =.00 M.00

Name: 5 1. Hdrogen and iodine gases react to form hdrogen iodide gas. If 6.00 mol of H and.00 mol of I are placed in a.00 L vessel and allowed to come to uilibrium at 50 C calculate the uilibrium concentrations of all species. The for the reaction is.00 at 50 C. H (g + I (g HI (g =.00 MR 1 1 I.00 1.00 0 C - - + E.00-1.00-.00.00(.00 8.00 8.00 I [ I (.00 1.00 1.00 0.667M Therefore [H =1. M, [I =0.0 M and [HI=1. M. ( (1.00.00 1. At 75 C, the uilibrium constant for the reaction between hdrogen and iodine gases to make hdrogen iodide gas is 51.5. A sample of hdrogen iodide was placed into a.00 L vessel and it was found that at uilibrium 0.18 mol of H gas was present. a How man moles of HI were originall placed into the flask? H (g + I (g HI (g = 51.5 V=.00 L MR 1 1 I 0 0 C + + - E - At uilibrium [H =0.109 M = Therefore at uilibrium [I =0.109 M and [HI=-0.18 I n [ I HI =(.00L(1.00 M =.00 mol ( 0.18 51.5 7.18 0.78 1.00/ < 1000 appro will not work 1.00 (0.109(0.109 0.18 0.109 0.18 b What are the uilibrium concentrations of I and HI? [I = [HI= 1.00-0.18 [I = 0.109 M [HI = 0.78 M Therefore.00 moles of hdrogen iodide were initiall added.

Name: 6 15. Carbon monoide and chlorine gases react to form phosgene gas (COCl. At uilibrium, there is 0.6 mol of CO, 0.8 mol of Cl and 1.56 mol of COCl in a.00 L container. How man moles of Cl must be added to reduce the concentration of CO to 0.5 mol/l? [0. mol CO (g + Cl (g COCl (g A Use the uilibrium concentrations given to calculate the value. [ COCl [ CO[ Cl (0.78 (0.1(0.1 18 B A given amount of chlorine gas in mol/l is added to the reaction and uilibrium is re-established at the SAME temperature. CO (g + Cl (g COCl (g =18 V=.0 L MR 1 1 1 I 0.1 0.1 + 0.78 0.1/ < 1000 C - - + appro will not E 0.1-0.1+- 0.78+ work At uilibrium [CO=0.5 M 0.1-=0.5 =0.06 M Therefore [Cl =0.1+-0.06 [COCl =0.78+0.06 = 0.08+ = 0.8 18 0.6.5 [ COCl [ CO[ Cl (0.8 (0.5(0.08.5 0.8 Therefore the moles of Cl added is 0. mol. 0.8 0.11 [Cl added = 0.11 mol/l 16. At 50 C the uilibrium constant for the reaction CaCO (s CaO (s + CO (g was found to be 0.15. a What is the uilibrium concentration of carbon dioide at this temperature? CaCO (s CaO (s + CO (g =[CO [CO =0.15 M b If 100.0 g of solid calcium carbonate was placed in a 10.0 L vessel and heated to the same temperature could uilibrium be reached? Justif our answer with calculations. Equilibrium will be reached with these conditions. 100.0 g of CaCO in 10.0 L moles of CaCO =1.00 mol Since 1 mol of CaCO produces 1 mol of CO, the maimum amount of carbon dioide that can be produced is 1.00 mol giving a [CO of 0.10 M. Since the concentration of CO when uilibrium is reached at this temperature is 0.15 M and the maimum [CO in this case is 0.10 M uilibrium will not be established.

Name: 7 17. The uilibrium constant for the reaction below is 0.11. Calculate all uilibrium concentrations if 0. mol of iodine chloride gas is placed in a 1.00 L vessel and allowed to come to uilibrium. ICl (g I (g + Cl (g =0.11 MR 1 1 I 0. 0 0 C - + + E 0.- 0./ < 1000 appro will not work ICl (g I (g + Cl (g 0.11 0. 0.11 0.11 [ I [ Cl [ ICl ( ( (0. 0. 0.66 1.66 0.066 Therefore [ICl =0.0 M, [I =0.066 M and [Cl =0.066 M. 18. At 100 C, the value of for the formation of hdrogen chloride gas from its elements is.5110. Determine the uilibrium concentrations of all species if 0.50 mol of both chlorine gas and hdrogen gas are placed in a 1.00 L vessel and allowed to reach uilibrium. H (g + Cl (g HCl (g Cl =.5110 [ Cl MR 1 1 I 0.50 0.50 0 (.5110 C - - + (0.50 (0.50 E 0.50-0.50-158 0.50/ < 1000 appro will not work 9.5 9.5 0.7 0.50 158 160 Therefore [H =.0010 - M, [Cl =.0010 - M and [HCl =0.9 M 19. Nitrogen gas reacts with ogen gas to form nitrogen monoide gas. The for this reaction at 150 C is 1.010 -. Determine the uilibrium concentrations of all species when 1.00 mol/l of nitrogen and ogen is placed in a vessel and allowed to reach uilibrium. N (g + O (g NO (g = 1.010 - MR 1 1 I 1.0 1.0 0 C - - + E 1.0-1.0-1.0/ >1000 appro will work 1.010 1.010 [ NO [ N [ O 5.810 (1.0 Therefore [N =0.995 M, [O =0.995 M and [NO =0.0110 M. ( (1.0

Name: 8 0. At 000 C, the for the decomposition of carbon dioide gas to carbon monoide and ogen gases is 6.010-7. Determine the ogen concentration at uilibrium when 1.00 mol of CO is placed in a 1.00 L vessel. CO (g CO (g + O (g MR 1 I 1.00 0 0 C - + + E 1.00-1.00/ >1000 appro will work Therefore the [O = 5.10 - M. 6.010 6.010 [ CO [ O [ CO 7 7 5.10 ( (1.00 1. The dissociation of ammonia gas to nitrogen and hdrogen gases has a value of.610-9 at 7 C. If.00 mol of ammonia is placed in a.00 L vessel and allowed to reach uilibrium, what is the concentration of hdrogen and nitrogen. NH (g N (g + H (g =.610-9 MR 1 I 1.00 0 0 C - + + E 1.00-1.00/ >1000 appro will work [ N.610.610 9.710 9 9 [ NH 11.110 Therefore [N =.110 M and [H =9.10 - M.. The following reaction takes place in a 1.00L vessel at 500 C. 7 ( ( ( (1.00 HI (g H (g + I (g Equilibrium concentrations were found to be 1.76 mol/l HI, 0.00 mol/l H and 0.00 mol/l I. If an additional 0.500 mol of hdrogen iodide gas is introduced at the same temperature, what the new concentrations of all gases once uilibrium has been reestablished? A Use the given concentrations to calculate [ I I (0.0(0.0 (1.76 1.010 B After the addition of an 0.500 mol of HI (g. HI (g H (g + I (g =1.010 - MR 1 1 I.6 0.00 0.00 C - + + E.6-0.00+ 0.00+ 0.00/ <1000 appro will not work 1.910 0.11 0.58 0.0580 I 0.07 Therefore [HI=.18 M, [H =0.1 M and [I =0.1 M. [ I (0.00 (0.00 (.6 0.00.6 0.8 0.00 1.8

Name: 9. The dissociation of ammonia to nitrogen and hdrogen gases at 00.0 C has a value of 1.9. If 0.500 mol of ammonia is placed in a 500.0 ml container, determine the uilibrium concentrations of all gases. NH (g N (g + H (g =1.9 [ N MR 1 [ NH I 1.00 0 0 C - + + ( ( 1.9 E 1.00- (1.00 1.00/ <1000 appro will not work If =-0.89 the concentrations of N and H will be negative which is not possible. 1.00/ <1000 appro will not work.00/ <1000 appro will not work 1.9 1.9 1.9 a 1 7 (1.00 5.0 1.00.78 5.0 5.0, b 0.89, 0.15 Therefore [NH =0.7 M, [N =0.15 M and [H =0.95 M..78, c 0.15. The uilibrium constant, for the reaction below is 85.0 at 60 C. If a miture is prepared where the initial concentration of sulfur dioide is 1.00 mol/l and nitrogen dioide is.00 mol/l calculate the uilibrium concentration of nitrogen monoide and nitrogen dioide. SO (g + NO (g NO (g + SO (g =85.0 MR 1 1 1 1 I 1.00.00 0 0 C - - + + E 1.00-.00-85.0 85.0(.00 170 Therefore [NO=0.989 M and [NO =1.01 M. a 1 [ NO[ SO [ SO (1.00 55 8.0, b.05, 0.989 [ NO.00 8.0 55, c 0 ( ( (.00 0.989 5. Hdrogen reacts with iodine vapour to produce hdrogen iodide vapour. The value of eg is 9.0.Calculate the concentration of all species at uilibrium, if there was.00 mol of hdrogen and 0.500 mol of iodine in a one vessel initiall. H (g + I (g HI (g = 9.0 I MR 1 1 [ I I.00 0.500 0 ( C - - + 9.0 (.00 (0.500 E.00-0.500-9.0(1.00 9.0 1 1.5 5, b., 0.87.50 Therefore [H =1.51 M, [I =0.010M and [HI=0.97 M. a 5 1.5, c 0.87 0 5 1.9 0 170

Name: 10 6. In a 1 L vessel,.0 g of phosphorus pentachloride is heated. Onl, 0 % of the phosphorus pentachloride dissociates into phosphorus trichloride and chlorine gas. What is the densit of the uilibrium miture? -.0 g/l - The Law of Conservation of Mass states that in a closed sstem mass is not lost during a chemical reaction therefore the densit of the uilibrium MIXTURE will be the same as the densit of the starting materials. 7. Phosphorus pentachloride decomposes into phosphorus trichloride and chlorine gas. If 1.9 % of a 1.0 mol/l concentration of phosphorus pentachloride decomposes, find the value of. PCl 5(g PCl (g + Cl (g MR 1 1 1 I 1.0 0 0 C - + + E 1.0- At uilibrium, 1.9% of PCl 5 is decomposed. 1.9% of 1.00 0.19 [ PCl 5 0.861M [ PCl [ PCl 0.19M Therefore the value is. 10 -. [ Cl 0.19M 8. The substance AO is 10.0 % molar dissociated according to the following reaction. [ PCl [ Cl (0.19(0.19 (0.861 5.10 AO (g A O (g + O (g Find the uilibrium concentration of each species if.0 mol of A O and 1.0 mol of O are initiall present in a 1.0 L volume. AO (g A O (g + O (g MR 1 I 0.0 1.0 C + - - E.0-1.0- At uilibrium, AO is 10% dissociated. [ A O 10% of.0 [ A O 0.0 0.90 Equil Equil.0 0.0 Therefore at uilibrium [AO =.6 M, [A O =0.0 M and [O =0.10 M.

Name: 11 9. If 0.70 mol of A, 0.50 mol of B, 0.0 mol of C and 0.90 mol of D were placed in 1.0 L container and allowed to come to uilibrium according to the following uation. A (g + B (g C (g + D (g = 0.500 Calculate the uilibrium concentrations of all species. A (g + B (g C (g + D (g = 0.500 MR 1 1 1 1 I 0.700 0.500 0.00 0.900 C - - + + E 0.700-0.500-0.00+ 0.900+ 0.00/ <1000 appro will not work 0.500 0.50 0.500(0.50 0.185 [ C[ D [ A[ B (0.00 (0.700 (0.60 (0.50 1.90 0.500, b 0.100 1.0 1.0 1.0 0.500 1.9, c (0.900 (0.500.70 0.185 0.100 Therefore at uilibrium [A=0.80 M, [B=0.60 M, [C=0.0 M and [D=0.80 M. a 1 0 (0.60 0. An uimolar concentration of A and B plus.00 mol of both C and D were placed in a 1.00 L vessel. A (g + B (g C (g + D (g = 6 Given that the uilibrium concentration of A is 1.00 M, calculate: a The uilibrium concentrations of all species. A (g + B (g C (g + D (g = 6 MR 1 1 1 I.00.00 C - - + + E - -.00+.00+ At uilibrium [A =1.00 1.00 [ B [ B 1.00 (1.00 0.667 [ B 1.00 Therefore at uilibrium [A=1.00 M, [B=0. M, [C=.67 M and [D=.67M. b The initial concentrations of A and B. 1.00 1.00 1.67 0.667 6 8 8 6 (.00 (.00 (1.00(1.00.00 1.00 8.00 9 [ C[ D [ A[ B Therefore the initial concentrations of A and B is 1.67 M. 1.0

Name: 1 1. The value for the formation of sulphur trioide gas from sulphur dioide and ogen gas is.5. If the uilibrium concentration of ogen was.0 M, determine the number of moles of sulfur trioide originall present in a 1.0 L vessel. [16 mol SO (g + O (g SO (g =.5 V=1.0L MR 1 I 0 0 C + + - E - At uilibrium [O =.0 M Therefore =.0 M [ SO.5 (.5( 1 1 16 [ SO (.0 ( ( ( [ O (.0.0 Therefore 16 moles of sulphur trioide were originall present in the vessel.