Asked by Clare
A 10.0-g sample of solid NH4Cl is heated in a 5.00-L container to 900.°C. At equilibrium the pressure of NH3(g) is 1.51 atm.
NH4Cl(s) mc011-1.jpg NH3(g) + HCl(g)
The equilibrium constant, Kp, for the reaction is:
NH4Cl(s) mc011-1.jpg NH3(g) + HCl(g)
The equilibrium constant, Kp, for the reaction is:
Answers
Answered by
DrBob222
I don't get all of the mumbo jumbo at the end of the question. I assume you want this.
..........NH4Cl --> NH3 + HCl
If pNH3 is 1.51 atm then pHCl is 1.51 atm.
Plug These numbers into the Kp expression and solve.
Kp = pNH3*pHCl
Kp = (1.51)(1.51) = ?
Note that NH4Cl doesn't appear in the Kp expression because it is a solid. Also note that the 10 g and 5.00 L container is just extraneous information. The concentration of NH4Cl, if you want it (but you don't need it) is 1.00.
..........NH4Cl --> NH3 + HCl
If pNH3 is 1.51 atm then pHCl is 1.51 atm.
Plug These numbers into the Kp expression and solve.
Kp = pNH3*pHCl
Kp = (1.51)(1.51) = ?
Note that NH4Cl doesn't appear in the Kp expression because it is a solid. Also note that the 10 g and 5.00 L container is just extraneous information. The concentration of NH4Cl, if you want it (but you don't need it) is 1.00.
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