Asked by Myles Tan
A 0.960mol quantity of Br2 is added to a 1.00 L reaction vessel that contains 1.23mol of H2 gas at 1000 K. What are the partial pressures of H2, Br2, and HBr at equilibrium?
At 1000 K, Kp=2.1×106 and ΔH∘ = -101.7 kJ for the reaction H2(g)+Br2(g)⇌2HBr(g).
At 1000 K, Kp=2.1×106 and ΔH∘ = -101.7 kJ for the reaction H2(g)+Br2(g)⇌2HBr(g).
Answers
Answered by
DrBob222
Use PV = nRT to solve for pH2 initial.
Use PV = nRT to solve for p
I obtained approx 100 for pBr2 and 128 for pH2 initially but you need to go through and do it more accurately.
...........H2 + B2 ==> 2HBr
I..........128..100.....0
C...........-p..-p......+2p
E.......128-p..100-p....+2p
Substitute the E line into the Kp expression and solve for p followed by working out partial pressures of each. Post your work if you run into trouble.
Use PV = nRT to solve for p
I obtained approx 100 for pBr2 and 128 for pH2 initially but you need to go through and do it more accurately.
...........H2 + B2 ==> 2HBr
I..........128..100.....0
C...........-p..-p......+2p
E.......128-p..100-p....+2p
Substitute the E line into the Kp expression and solve for p followed by working out partial pressures of each. Post your work if you run into trouble.
Answered by
Myles Tan
How did you get 100 and 128? I got 101 and 78.
Answered by
Anonymous
A 0.960mol quantity of Br2 is added to a 1.00 L reaction vessel that contains 1.23mol of H2 gas at 1000 K. What are the partial pressures of H2, Br2, and HBr at equilibrium?
At 1000 K, Kp=2.1×106 and ΔH∘ = -101.7 kJ for the reaction H2(g)+Br2(g)⇌2HBr(g).
Dr Bob, you said:
Use PV = nRT to solve for pH2 initial.
Use PV = nRT to solve for p
I obtained approx 100 for pBr2 and 128 for pH2 initially but you need to go through and do it more accurately.
...........H2 + B2 ==> 2HBr
I..........128..100.....0
C...........-p..-p......+2p
E.......128-p..100-p....+2p
How did you get 100 and 128? I got 101 and 78.
At 1000 K, Kp=2.1×106 and ΔH∘ = -101.7 kJ for the reaction H2(g)+Br2(g)⇌2HBr(g).
Dr Bob, you said:
Use PV = nRT to solve for pH2 initial.
Use PV = nRT to solve for p
I obtained approx 100 for pBr2 and 128 for pH2 initially but you need to go through and do it more accurately.
...........H2 + B2 ==> 2HBr
I..........128..100.....0
C...........-p..-p......+2p
E.......128-p..100-p....+2p
How did you get 100 and 128? I got 101 and 78.
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