Addition of Silver nitrate water solution to weak solution of Potassium chromate leads to precipitation in case of: C'(Ag')·C(CrO,) > K,°(Ag,CrO4) C(AGNO3) > C(K,CrO4) C(Ag') = C(CrO,²) C(Ag')·C(CrO,²)= K,°(Ag,CrO4)
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- Describe a nonchemical system that is not in equilibrium, and explain why equilibrium has not been achieved.Consider a metal ion A2+ and its nitrate salt, In an experiment, 35.00 mL of a 0.217 M solution of A(NO3)2 is made to react with 25.00 mL of 0.195 M NaOH. A precipitate, A(OH)2, forms. Along with the precipitation, the temperature increases from 24.8C to 28.2C. What is H for the precipitation of A(OH)2? The following assumptions can be made. • The density of the solution is 1.00 g/mL. • Volumes are additive. • The specific heat of the solution is 4.18 J/g C.When 7.11 g NH4NO3 is added to 100 mL water, the temperature of the calorimeter contents decreases from 22.1 C to 17.1 C. Assuming that the mixture has the same specific heat as water and a mass of 107 g, calculate the heat q. Is the dissolution of ammonium nitrate exothermic or endothermic?
- Addition of Bismuth nitrate water solution to weak solution of Potassium iodide leads to precipitation in case of: C (Bi ³*)· C'(J)>K, (Bi J;) C(Bỉ")' ·C (J)>K,°( BiJ;) - 3+3 C( Bi )·C(J) = K,°( BiJ;) C (Bỉ*)-C*(J)Addition of Barium nitrate water solution to weak solution of Potassium phosphate leads to precipitation in case of: C'(Ba")-C°(PO,¬>K°(Ba;(PO4)2) C(Ba(NO3)2) > C(K;PO4) C(Ba")·C(PO,) = K,°(Ba;(PO4)2) C'(Ba")-C(PO,')Potassium chromate and lead (II) acetate are both dissolved in a beaker of water, where they react to form solid lead (II) chromate. What is the balanced equation describing this reaction? K:Cro.(aq) + Pb(C-H:O-):(sq) → PbCrO«(,) + KC:H:0:(3q) O 2K:Cro.(aq) + Pb(C:H:O-):(sq) PbCrO.() + 2KC:H.0:(aq) O K:Cro.(aq) + Pb(C:H.O:):(aq) -> PbCrO.(.) + 2KC:H:0:(aq) 2K.CrO.(sq) + Pb(C:H:0:):(aq) - PbCrO-(.) + KC:H:0:(aq) -->3.2 The standard Gibbs energy of formation of H20 at 25.0 °C, 4G, is -228.57 kJ.mot', for a standard state of 1 atm. The reaction is: 2 H30 (g) - O (g) + 2H, (g) (a) Calculate 4 G° (kJ.mol) for the above reaction. (b) Calculate the value of the equilibrium constant, K, for the above reaction. (c) Calculate the value of K at 5 atm.Silver chloride is classified as an insoluble compound because it dissolves only to a small extent in water. Consider the balanced equation and equilibrium constant that shows the partial dissolving of silver chloride in water: AgCl(s) = Ag*(aq) + CI"(ag) Keg = 1.70 x 10-10 a) The reaction AgCl(s) = Ag*(aq) + Cl (ag) is classified as a (homogeneous or heterogeneous) equilibrium because it involves more than one physical state, solid and aqueous. b) Based on the equilibrium constant value and the equilibrium constant expression, the concentration of each ion in a saturated solution of silver chloride is : Concentration of Ag = x 10 Concentration of CI" = x 10 MH3C. (k) Zn/CH;COOH > 1Q + 1R heat H3C2. In the determination of water of crystallization for the hydrated salt BaCl2.2H20, the following data were obtained by a student. = 22.3505 g = 26.0005 Mass of crucible and lid (g) Mass of crucible, lid and sample (g) Mass of crucible, lid and sample after 1st heating (g) = 25.4931 Mass of crucible, lid and sample after 1$st heating (g) = 25.4931 a) What is the mass of the BaCl2.2H2O sample? Solution: b) What is the mass of water driven off? Solution:C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l)ΔH = –2802.7 kJ mol –1a) Write an expression for the equilibrium constant for this reaction.b) At equilibrium, the concentration of the reactants and products are determined as [CO2] = 0.30 M, [O2] = 0.040 M and [C6H12O6] = 0.065 M. Determine the value of the equilibrium constant (Kc) and predict the whether the products or reactants will be favoured at equilibrium.c) Given that the concentrations of the reactants and products at a particular time are [CO2] = 0.65 M, [O2] = 0.020 M and [C6H12O6] = 0.055 M, determine the reaction quotient (Qc). Compare the Kc and Qc values and predict the favoured direction of the reaction.d) Explain the effect on equilibrium of:i) Increasing temperatureii) Increasing pressureiii) Decreasing the concentration of oxygeniv) Increasing the concentration of carbon dioxidev) Adding a catalyst3.2 The standard Gibbs energy of formation of H20 at 25.0 °C, A,G°, is -228.57 kJ.mol', for a standard state of 1 atm. The reaction is: 2 H20 (g)→02 (g) +2H2 (g) (a) Calculate 4,G° (kJ.mol') for the above reaction. (b) Calculate the value of the equilibrium constant, K, for the above reaction. (c) Calculate the value of K, at 5 atm.SEE MORE QUESTIONSRecommended textbooks for youChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
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