1
answer
0
watching
58
views
19 Nov 2019
consider the following set of equilibrium reactions with the corresponding... help!! 1. Consider the flowing set of equilibrium reactions with the corresponding equilibrium constants KI CHM Cg) H20 (g) COCg) 3H20g) K 1.930 x 10 CO(g) H200g) CO2 (g) H2(g) K2 5.528 Initially, a gas containing 20% CH4 and 80% H20 is present at 2000k & P-latm. Choose a basis of 10 moles of gas initially present and let el cqual the degrec of reaction for the first reaction and e2 equal the degree of reaction for the second reaction. The equilibrium constants for the first and the second gas phase reaction can be written as CO H CO2 rH2 CH. H20 co H20 The compositions present in the equilibrium expressions can be written in terms of the degree of reaction using the stoichiometry of the reactions. For example, for the mole fraction of CO, CO is formed by the first reaction and consumed by the second (i.e., e e2). The mole fraction of Co is 1- then (e e2) divided by the total number of moles (10+2e1) after reaction. 3e, e. 8 e1 -e2 2 e, e1 e2 CO 10 2e. H2 10 2e. H20 CO2 CHA 10 2e 10 2e. 10 2e Substituting these expressions into the equilibrium constants and rearrangement yields: (e1 2)(3e1 e2) 1.930 x 10 eu )(8 e e2)(10 2e1) e2(3e1 e2) 5.528 008 e1 e2) Solve for the equilibrium gas composition for this system Hint: Remember the discussion on "scaling" at the end of Lecture 6
consider the following set of equilibrium reactions with the corresponding... help!!
1. Consider the flowing set of equilibrium reactions with the corresponding equilibrium constants KI CHM Cg) H20 (g) COCg) 3H20g) K 1.930 x 10 CO(g) H200g) CO2 (g) H2(g) K2 5.528 Initially, a gas containing 20% CH4 and 80% H20 is present at 2000k & P-latm. Choose a basis of 10 moles of gas initially present and let el cqual the degrec of reaction for the first reaction and e2 equal the degree of reaction for the second reaction. The equilibrium constants for the first and the second gas phase reaction can be written as CO H CO2 rH2 CH. H20 co H20 The compositions present in the equilibrium expressions can be written in terms of the degree of reaction using the stoichiometry of the reactions. For example, for the mole fraction of CO, CO is formed by the first reaction and consumed by the second (i.e., e e2). The mole fraction of Co is 1- then (e e2) divided by the total number of moles (10+2e1) after reaction. 3e, e. 8 e1 -e2 2 e, e1 e2 CO 10 2e. H2 10 2e. H20 CO2 CHA 10 2e 10 2e. 10 2e Substituting these expressions into the equilibrium constants and rearrangement yields: (e1 2)(3e1 e2) 1.930 x 10 eu )(8 e e2)(10 2e1) e2(3e1 e2) 5.528 008 e1 e2) Solve for the equilibrium gas composition for this system Hint: Remember the discussion on "scaling" at the end of Lecture 6
Nelly StrackeLv2
21 Sep 2019