The system
H2(g) + I2(g) ⇌ 2HI(g)
is at equilibrium at a fixed temperature with a partial pressure of H2 of 0.200 atm, a partial pressure of I2 of 0.200 atm, and a partial pressure of HI of 0.100 atm. An additional 0.26 atm pressure of HI is admitted to the container, and it is allowed to come to equilibrium again. What is the new partial pressure of HI?
0.464 atm
0.104 atm
0.360 atm
0.152 atm
PARTIAL PRESSURE CHEMISTRY QUESTIOON? :(?
2016-02-27 5:32 pm
回答 (2)
2016-02-27 5:56 pm
H₂(g) + I₂(g) ⇌ 2HI(g) ..... Kp = P(HI)²/{P(H₂)×P(I₂)}
In the original equilibrium state:
P(H₂) = P(I₂) = 0.200 atm
P(HI) = 0.100 atm
In the new equilibrium state :
P(H₂) = P(I₂) = (0.200 + y) atm
P(HI) = (0.100 + 0.260 - y) atm = (0.360 - y) atm
At fixed temperature, equilibrium constant is unchanged.
(0.360 - y)²/(0.200 + y)² = (0.100)²/(0.200)²
(0.360 - y)/(0.200 + y) = (0.100)/(0.200)
(0.360 - y)/(0.200 + y) = 1/2
0.200 + y = 2(0.360 - y)
0.200 + y = 0.720 - 2y
3y = 0.520
y = 0.173
new P(HI) = (0.360 - 0.173) = 0.187
...... There is no correct answer in the given options.
In the original equilibrium state:
P(H₂) = P(I₂) = 0.200 atm
P(HI) = 0.100 atm
In the new equilibrium state :
P(H₂) = P(I₂) = (0.200 + y) atm
P(HI) = (0.100 + 0.260 - y) atm = (0.360 - y) atm
At fixed temperature, equilibrium constant is unchanged.
(0.360 - y)²/(0.200 + y)² = (0.100)²/(0.200)²
(0.360 - y)/(0.200 + y) = (0.100)/(0.200)
(0.360 - y)/(0.200 + y) = 1/2
0.200 + y = 2(0.360 - y)
0.200 + y = 0.720 - 2y
3y = 0.520
y = 0.173
new P(HI) = (0.360 - 0.173) = 0.187
...... There is no correct answer in the given options.
2016-02-27 5:59 pm
K = (pHI)^2/pH2(pI2)
K = (0.10)^/0.20(0.20
K = 0.25
pHI = 0.10 + 0.26 = 0.36
Let 2x = change in pHI
At equilibrium = pHI = 0.36-2x
Increase in pH2 = +x
Increase in pI2 = +x
At equilibrium
pI2 = 0.20 + x ; pH2 = 0.20 +x
(0.36-2x)^2/(0.20+x)(0.20 + x) = 0.25
Take square root of both sides:
(0.36 -2x)/0.20 + x ) = 0.50
0.36-2x = 0.50(0.20 + x)
0.36-2x = 0.10+ 0.50x
2.5x = 0.26
x = 0.104
partial pressure of HI = 0.36-2x = 0.36 -2(.104) = 0.152
K = (0.10)^/0.20(0.20
K = 0.25
pHI = 0.10 + 0.26 = 0.36
Let 2x = change in pHI
At equilibrium = pHI = 0.36-2x
Increase in pH2 = +x
Increase in pI2 = +x
At equilibrium
pI2 = 0.20 + x ; pH2 = 0.20 +x
(0.36-2x)^2/(0.20+x)(0.20 + x) = 0.25
Take square root of both sides:
(0.36 -2x)/0.20 + x ) = 0.50
0.36-2x = 0.50(0.20 + x)
0.36-2x = 0.10+ 0.50x
2.5x = 0.26
x = 0.104
partial pressure of HI = 0.36-2x = 0.36 -2(.104) = 0.152
收錄日期: 2021-04-18 14:38:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160227093233AAXEbnm