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A Level question on Kp
Hey
I got this question for homework, can anyone help me?
When 0.50 mol of iodine and 0.50 mol of hydrogen were mixed in a closed container at 723 K and 2 atm pressure, 0.11 mol of iodine were found to be present when equilibrium was established.
Calculate the partial pressures of iodine, hydrogen and HI in the equilibrium mixture.
H2 + I2 = 2HI (All gases)
Thankyou soooo much if you can help!
I got this question for homework, can anyone help me?
When 0.50 mol of iodine and 0.50 mol of hydrogen were mixed in a closed container at 723 K and 2 atm pressure, 0.11 mol of iodine were found to be present when equilibrium was established.
Calculate the partial pressures of iodine, hydrogen and HI in the equilibrium mixture.
H2 + I2 = 2HI (All gases)
Thankyou soooo much if you can help!
Reply 1
Kp= p(HI)²/p(H2)*p(I2)
at equilibrium: 0.11(H2) + 0.11(I2) <-> (0.78)(HI)
mole fraction: 0.11 0.11 0.78
partial pressure: 0.22 0.22 1.56
sub the pp into the equation
gives: 50.28 (no units as they cancel out)
at equilibrium: 0.11(H2) + 0.11(I2) <-> (0.78)(HI)
mole fraction: 0.11 0.11 0.78
partial pressure: 0.22 0.22 1.56
sub the pp into the equation
gives: 50.28 (no units as they cancel out)
Reply 2
CHAD
OP
gordon2002
Kp= p(HI)²/p(H2)*p(I2)
at equilibrium: 0.11(H2) + 0.11(I2) <-> (0.78)(HI)
mole fraction: 0.11 0.11 0.78
partial pressure: 0.22 0.22 1.56
sub the pp into the equation
gives: 50.28 (no units as they cancel out)
at equilibrium: 0.11(H2) + 0.11(I2) <-> (0.78)(HI)
mole fraction: 0.11 0.11 0.78
partial pressure: 0.22 0.22 1.56
sub the pp into the equation
gives: 50.28 (no units as they cancel out)
wicked, thanks!
Reply 3
no problem
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