Asked by Kyle
2ππ»3(π)β·π2(π)+3π»2(π) πΎπ=0.83
Consider your answers above, if the initial pressures for all three species is 1 atm what is the equilibrium pressure of H2? (Hint: Your quadratic will have two solutions, which one is impossible?)
Consider your answers above, if the initial pressures for all three species is 1 atm what is the equilibrium pressure of H2? (Hint: Your quadratic will have two solutions, which one is impossible?)
Answers
Answered by
DrBob222
I'm little out of the loop since I don't know your previous answers but here is the solution to this problem.
...........................2ππ»3(π)β·π2(π)+3π»2(π) πΎπ=0.83
I..............................1...............1..........1
C............................-2p.............p...........3p
E............................1-2p.........1+p.........1+3p
Kp = (N2)(H2)^3/(NH3)^2
0.83 = (1+p)(1+3p)^3/(1-2p)^2
Solve that. The chemistry part is abovel All that is left is the math part.
Post your work if you get stuck.
...........................2ππ»3(π)β·π2(π)+3π»2(π) πΎπ=0.83
I..............................1...............1..........1
C............................-2p.............p...........3p
E............................1-2p.........1+p.........1+3p
Kp = (N2)(H2)^3/(NH3)^2
0.83 = (1+p)(1+3p)^3/(1-2p)^2
Solve that. The chemistry part is abovel All that is left is the math part.
Post your work if you get stuck.
Answered by
Kyle
Yeah, I got the answer to be 0.9
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