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 CH 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. Classify 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 (0.15)(0.7) (0.010) 5 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) [ NO [ N O [ NO 0.1 (0.155) [ NO 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 (0.8) (0.70) [ NO [ N O - (5.910 ) (6.810 ) 5.910 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 (0.100 )(0.100 ) 9. (0.100 ) 0.9 10. 0.090 0.100/ < 100 appro will not work [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) 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 [ CO[ Cl [ COCl 10 ( )( ).10 0.100 10.10 0.100 6.710 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.00 [SO =.00 M
Name: 5 1. Hydrogen and iodine gases react to form hydrogen 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 1.00/ < 1000 appro will not work I [ I ().00 (.00 )(1.00 ).00(.00.00 ) 8.00 1.00 8.00 1.00 0.667M Therefore [H =1. M, [I =0.0 M and [HI=1. M. 1. At 75 C, the uilibrium constant for the reaction between hydrogen and iodine gases to make hydrogen iodide gas is 51.5. A sample of hydrogen 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 many moles of HI were originally placed into the flask? H (g) + I (g) HI (g) = 51.5 V=.00 L MR 1 1 I 0 0 y C + + E y At uilibrium [H =0.109 M = Therefore at uilibrium [I =0.109 M and [HI=y 0.18 I n [ I HI =(.00L)(1.00 M) =.00 mol ( y 0.18) 51.5 (0.109)(0.109) Therefore.00 moles of hydrogen iodide were initially added. y 0.18 7.18 0.109 0.78 y 0.18 y 1.00 b) What are the uilibrium concentrations of I and HI? [I = [HI= 1.00 0.18 [I = 0.109 M [HI = 0.78 M
Name: 6 15. 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? Justify your 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. 16. 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) ICl (g) I (g) + Cl (g) =0.11 MR 1 1 I 0. 0 0 C + + E 0. 0./ < 1000 appro will not work [ I [ Cl [ ICl ( )( ) 0.11 (0. ) 0. 0. 0.11 0.66 0.11 1.66 0.066 Therefore [ICl =0.0 M, [I =0.066 M and [Cl =0.066 M. 17. At 100 C, the value of for the formation of hydrogen chloride gas from its elements is.5110. Determine the uilibrium concentrations of all species if 0.50 mol of both chlorine gas and hydrogen 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 0.50/ < 1000 9.5 158 appro will not work 9.5 160 0.7 Therefore [H =.0010 M, [Cl =.0010 M and [HCl =0.9 M
Name: 7 18. Nitrogen gas reacts with oygen 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 oygen is placed in a vessel and allowed to reach uilibrium. N (g) + O (g) NO (g) = 1.010 [ NO MR 1 1 [ N [ O I 1.0 1.0 0 () C + 1.010 E 1.0 1.0 (1.0 )(1.0 ) 1.010 1.0/ >100 5.810 appro will work Therefore [N =0.995 M, [O =0.995 M and [NO =0.0110 M. 19. At 000 C, the for the decomposition of carbon dioide gas to carbon monoide and oygen gases is 6.010 7. Determine the oygen concentration at uilibrium when 1.00 mol of CO is placed in a 1.00 L vessel. CO (g) CO (g) + O (g) [ CO [ O MR 1 [ CO I 1.00 0 0 7 () ( ) C + + 6.010 E 1.00 (1.00 ) 1.00/ >100 appro will work Therefore the [O = 5.10 M. 6.010 7 5.10 0. The dissociation of ammonia gas to nitrogen and hydrogen 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 hydrogen and nitrogen. NH (g) N (g) + H (g) =.610 9 [ N MR 1 [ NH I 1.00 0 0 C + + 9 ( )().610 E 1.00 (1.00 ) 1.00/ >100 appro will work.610 9.710 9 11.110 Therefore [N =.110 M and [H =9.10 M. 7 1. The following reaction takes place in a 1.00L vessel at 500 C. 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 hydrogen 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
Name: 8 [ I I [ I (0.0)(0.0) I (1.76) (0.00 )(0.00 ) 1.010 1.910 (.6 ) B) After the addition of an 0.500 mol of HI (g). 0.00 HI (g) H (g) + I (g) =1.010 0.11.6 MR 1 1 I.6 0.00 0.00 0.58 0.8 0.00 C + + 0.0580 1.8 E.6 0.00+ 0.00+ 0. 07 0.00/ <100 appro will not work Therefore [HI=.18 M, [H =0.1 M and [I =0.1 M.. The dissociation of ammonia to nitrogen and hydrogen 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/ <100 7 1.9 appro will not work (1.00 ) If = 0.89 the concentrations of N and H will be negative which is not possible. 5.0 1.9 1.00 1.9.78 5.0 a 5.0, b.78, c 1.9 0.89, 0.15 0.15 Therefore [NH =0.7 M, [N =0.15 M and [H =0.95 M.. 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 [ NO[ SO MR 1 1 1 1 [ SO [ NO I 1.00.00 0 0 ( )( ) C + + 85.0 (1.00 )(.00 ) E 1.00.00 85.0(.00.00 ) 1.00/ <100 appro will not work 170 55 8.0 0 a 8.0, b 55, c 170 0.989 Therefore [NO=0.989 M and [NO =1.01 M. 1 1.05, 0 0.989
Name: 9. Hydrogen reacts with iodine vapour to produce hydrogen iodide vapour. The value of eg is 9.0.Calculate the concentration of all species at uilibrium, if there was.00 mol of hydrogen and 0.500 mol of iodine in a one vessel initially. 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 0.500.00/ <100 appro will not work 9.0(1.00.50 9.0 1.5 5 a 5, b 1.5, c 5., 0.87 Therefore [H =1.51 M, [I =0.010M and [HI=0.97 M. 1 0.87 ) 0 5. 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 0.861M 5 [ PCl [ Cl [ PCl (0.19)(0.19) (0.861) [ PCl 0.19M Therefore the value is. 10. [ Cl 0.19M 6. The substance AO is 10.0 % molar dissociated according to the following reaction. 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 initially 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 Equil 0.0.0 0.90 Equil 0.0 5.10 Therefore at uilibrium [AO =.6 M, [A O =0.0 M and [O =0.10 M.
Name: 10 7. 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/ <100 appro will not work [ C[ D [ A[ B (0.00 )(0.900 ) 0.500 (0.700 )(0.500 ) (0.60 1.0 0.50 (0.50 1.0 0.500(0.50 1.0 0.185 1.90 0.500 a 0.500, b 1.9, c 0.185 0.100.70 0 0.100 Therefore at uilibrium [A=0.80 M, [B=0.60 M, [C=0.0 M and [D=0.80 M. 1 ) ) ) (0.60 1.0 8. 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 y y.00.00 C + + E y y.00+.00+ At uilibrium [A =1.00 y 1.00 y 1.00 [ B y [ B (1.00 ) (.00 )(.00 ) 6 (1.00)(1.00 ).00 8 1.00 8 8.00 6 9 [ C[ D [ A[ B 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. y 1.00 y 1.00 0.667 y 1.67 Therefore the initial concentrations of A and B is 1.67 M. )
Name: 11 9. The value for the formation of sulphur trioide gas from sulphur dioide and oygen gas is.5. If the uilibrium concentration of oygen was.0 M, determine the number of moles of sulfur trioide originally present in a 1.0 L vessel. [16 mol SO (g) + O (g) SO (g) =.5 V=1.0L MR 1 I 0 0 y C + + E y At uilibrium [O =.0 M Therefore =.0 M [ SO [ SO [ O ( y.0).5 () () (.5)() ( y.0) 1 ( y.0) 1 y y 16 Therefore 16 moles of sulphur trioide were originally present in the vessel.
Name: 1 0. 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 many 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 y 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 [ COCl I 0.1 0.1 +y 0.78 0.1/ < 1000 [ CO[ Cl C + appro will not E 0.1 0.1+y 0.78+ work (0.8) 18 At uilibrium [CO=0.5 M (0.5)(0.08 y) 0.1 =0.5 =0.06 M Therefore [Cl =0.1+y 0.06 [COCl =0.78+0.06 = 0.08+y = 0.8 0.6.5y 0.8.5y 0.8 y 0.11 [Cl added = 0.11 mol/l Therefore the moles of Cl added is 0. mol.