78. What are the concentrations of PCI5, PCI3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of pure PCI5 with [PCI5] = 2.00 M? %3D PCis(g) = PC13(g) + Cl2(g) K = 0.0211

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What are the concentrations of PCL5,PCL3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of pure PCL5 with [PCL5] = 2.00M PCL5(g) = PCL3(g) + Cl2(g) Kc = 0.0211
he change is small enough to be neglected.
75. Assume that the change in concentration of COCI2 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCI,
with an initial concentration of 0.3166 M.
COC2(g) = C0(g) + Cl2(g)
K. = 2.2 × 10-10
(b) Confirm that the change is small enough to be neglected.
76. Assume that the change in pressure of H2S is small enough to be neglected in the following problem.
(a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with
an initial pressure of 0.824 atm.
2H2S(g) = 2H2(g) + S2(g)
Kp = 2.2 x 10-6
(b) Confirm that the change is small enough to be neglected.
77. What are all concentrations after a mixture that contains [H2O] = 1.00 M and [Cl20] = 1.00 M comes to equilibrium at 25 °C?
H2O(g) + Cl½0(g) = 2HOCI(g)
K. = 0.0900
78. What are the concentrations of PCI5, PC|3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of
pure PCI5 with [PC\5] = 2.00 M?
PCİ5(g) = PC13(g) + Cl2(g)
K. = 0.0211
79. Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H2 and 63.5 g of iodine at 448 °C.
H2 + I2 = 2HI
K. = 50.2 at 448 °C
80. Butane exists as two isomers, n-butane and isobutane.
CH3
CH2
CH3
CH3
CH2
CH
CH3
`CH3
n-butane
isobutane
Kp = 2.5 at 25 °C
What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?
81. What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the
equilibrium constant (Ke) is 0.50 for the decomposition reaction of CaCO3 at that temperature?
CACO3(s) = CaO(s) + CO2 (g)
Transcribed Image Text:he change is small enough to be neglected. 75. Assume that the change in concentration of COCI2 is small enough to be neglected in the following problem. (a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCI, with an initial concentration of 0.3166 M. COC2(g) = C0(g) + Cl2(g) K. = 2.2 × 10-10 (b) Confirm that the change is small enough to be neglected. 76. Assume that the change in pressure of H2S is small enough to be neglected in the following problem. (a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H2S with an initial pressure of 0.824 atm. 2H2S(g) = 2H2(g) + S2(g) Kp = 2.2 x 10-6 (b) Confirm that the change is small enough to be neglected. 77. What are all concentrations after a mixture that contains [H2O] = 1.00 M and [Cl20] = 1.00 M comes to equilibrium at 25 °C? H2O(g) + Cl½0(g) = 2HOCI(g) K. = 0.0900 78. What are the concentrations of PCI5, PC|3, and Cl2 in an equilibrium mixture produced by the decomposition of a sample of pure PCI5 with [PC\5] = 2.00 M? PCİ5(g) = PC13(g) + Cl2(g) K. = 0.0211 79. Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H2 and 63.5 g of iodine at 448 °C. H2 + I2 = 2HI K. = 50.2 at 448 °C 80. Butane exists as two isomers, n-butane and isobutane. CH3 CH2 CH3 CH3 CH2 CH CH3 `CH3 n-butane isobutane Kp = 2.5 at 25 °C What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm? 81. What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (Ke) is 0.50 for the decomposition reaction of CaCO3 at that temperature? CACO3(s) = CaO(s) + CO2 (g)
Expert Solution
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At a certain temperature decomposition reaction of PCl5(g) forms PCl3(g) and Cl2(g) as products and make an equilibrium mixture. The decomposition reaction given below.

PCl5(g) = PCl3(g) + Cl2(g)

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