A mixture of 0.03538 mol of O2, 0.03149mol of NOCI, 0.06714 mol of NO2, and0.02565 mol of Cl2 is placed in a 1.0-L steelpressure vessel at 1377 K. The followingequilibrium is established: 1 02(g) + 2NOCI((g) 2 NO2(g) + 1 Cl2(g) At equilibrium0.01983 mol of C12 is found in the reactionmixture.(a) Calculate the equilibrium partialpressures of O2, NOCI, NO2, and C12.Peq(02) =Peq(NOCI) = .Peq(NO2) = .Peq(CI2) = .(b) Calculate KP for this reaction. KP =

#(a): We are given the initial concentration of each of the reactants and products of the…

mol of NOCI, 0.06714 mol of NO2, and 0.02565 mol of Cl2 is placed in a 1.0-L steel pressure vessel at 1377 K. The following equilibrium is established: 1 O2(g) + 2 NOCI(g) 2 NO2(g) + 1 Cl2(g) At equilibrium 0.01983 mol of Cl2 is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of 02, NOCI, NO2, and Cl2. Peq(02) = . Peq(NOCI) = . Peq(NO2) = . Peq(Cl2) = . (b) Calculate KP for this reaction. KP =

mol of NOCI, 0.06714 mol of NO2, and 0.02565 mol of Cl2 is placed in a 1.0-L steel pressure vessel at 1377 K. The following equilibrium is established: 1 O2(g) + 2 NOCI(g) 2 NO2(g) + 1 Cl2(g) At equilibrium 0.01983 mol of Cl2 is found in the reaction mixture. (a) Calculate the equilibrium partial pressures of 02, NOCI, NO2, and Cl2. Peq(02) = . Peq(NOCI) = . Peq(NO2) = . Peq(Cl2) = . (b) Calculate KP for this reaction. KP =

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 14.3QE: Describe a nonchemical system that is not in equilibrium, and explain why equilibrium has not been...

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Crystal Lattices and Unit Cell

Consider a crystal. The regular and repeating pattern of constituent particles is the most distinguishing feature of crystalline solids. A crystal lattice is formed when these particles are replaced with representative points. The three-dimensional arrangement of constituent particles is known as a crystal lattice. Unit cell is that the smallest portion of a space lattice which when repeated in several directions, generate the whole lattice. Unit cell is a building block of a lattice. It is the smallest unit of cell.

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A mixture of 0.03538 mol of O2, 0.03149
mol of NOCI, 0.06714 mol of NO2, and
0.02565 mol of Cl2 is placed in a 1.0-L steel
pressure vessel at 1377 K. The following
equilibrium is established: 1 02(g) + 2
NOCI((g) 2 NO2(g) + 1 Cl2(g) At equilibrium
0.01983 mol of C12 is found in the reaction
mixture.
(a) Calculate the equilibrium partial
pressures of O2, NOCI, NO2, and C12.
Peq(02) =
Peq(NOCI) = .
Peq(NO2) = .
Peq(CI2) = .
(b) Calculate KP for this reaction. KP =

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