Question

An equilibrium mixture for the following reaction: H2(g) + I2(g) <---> 2HI(g) is composed of the following: P(I2) = 0.08592 atm; P(H2) = 0.08592 atm; P(HI) = 0.5996 atm. If this equilibrium is disturbed by adding more HI so that the partial pressure of HI is suddenly increased to 1.0000 atm, what will the partial pressures of each of the gases be when the system returns to equilibrium?

Answer #1

H_{2}(g) + I_{2}(g) <---> 2HI(g)

Given:

P(I_{2}) = 0.08592 atm.

P(H_{2}) = 0.08592 atm.

P(HI) = 0.5996 atm.

Calculate Kp of reaction

Kp = [P(HI)]^2/ P(I_{2}) P(H_{2})

Kp = [0.5996]^2/([0.08592] [0.08592]) =
**48.7**

Now this equilibrium is disturbed by adding more HI so that the partial pressure of HI is suddenly increased to 1.0000 atm.

The partial pressures of each of the gases, when the system returns to equilibrium

Let the partial pressures of P(I_{2}) be x and
P(H_{2}) = x

Then at equilibrium

Kp = [P(HI)]^2/ P(I_{2}) P(H_{2})

48.7 = (1)^2/x*x

x^2 = (1)^2/ 48.7

x^2 = 0.020533

x = 0.14329 = P(I_{2}) = P(H_{2})

The partial pressures of each of the gases, when the system returns to equilibrium

**P(I _{2}) = 0.14329 atm.**

**P(H _{2}) = 0.14329 atm.**

**P(HI) = 1.0000 atm.**

Consider the following reaction:
H2(g)+I2(g)⇌2HI(g)
A reaction mixture at equilibrium at 175 K
containsPH2=0.958atm, PI2=0.877atm, and
PHI=0.020atm. A second reaction mixture, also at 175 K,
contains PH2=PI2= 0.628 atm , and PHI=
0.107 atm .
A) If not, what will be the partial pressure of HI when the
reaction reaches equilibrium at 175 K?

Consider the following reaction: H2(g)+I2(g)?2HI(g) A reaction
mixture at equilibrium at 175 K contains PH2=0.958atm,
PI2=0.877atm, and PHI=0.020atm. A second reaction mixture, also at
175 K, contains PH2=PI2= 0.630 atm , and PHI= 0.102 atm .
Is the second reaction at equilibrium?
If not, what will be the partial pressure of HI when the
reaction reaches equilibrium at 175 K?

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
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come to equilibrium again. What is the new partial pressure of
HI?
A.0.360 atm
B. 0.464 atm
C. 0.152 atm
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Consider the reaction:
H2(g)+I2(g)⇌2HI(g)
A reaction mixture at equilibrium at 175 K contains
PH2=0.958atm, PI2=0.877atm,
and PHI=0.020atm. A second reaction mixture,
also at 175 K, contains
PH2=PI2=0.629
atm , and PHI= 0.101 atm .
Is the second reaction at equillibrium? I found that the
kp=4.76X10-4 and Qp = .0257 So no, not at
equillibrium.
If not, what is the partial pressure of HI when the
reaction reaches equilibrium at 175 K? I need help
figuring out the ICE chart and...

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H2 and I2 is:
H2(g) + I2(g) <--> 2HI(g)
The value of Kp for the reaction is 69.0 at 730.0C.
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vessel at 730.0C if the initial partial pressures of H2
and I2 are both 0.1600 atm and initially there is no HI
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Consider the following reaction: H2(g)+I2(g)⇌2HI(g) A reaction
mixture in a 3.75 L flask at a certain temperature initially
contains 0.764 g H2 and 97.1 g I2. At equilibrium, the flask
contains 90.4 g HI. Calculate the equilibrium constant (Kc) for the
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Consider the following reaction: H2(g)+I2(g)⇌2HI(g) A reaction
mixture in a 3.63 L flask at a certain temperature initially
contains 0.767 g H2 and 97.0 g I2. At equilibrium, the flask
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significant figures.

Consider the following reaction:
H2(g)+I2(g)⇌2HI(g)
A reaction mixture at equilibrium at 175 Kcontains
PH2=0.958atm, PI2=0.877atm, and
PHI=0.020atm. A second reaction mixture, also at 175 K,
contains PH2=PI2=0.622 atm , and PHI=
0.109 atm

The equilibrium constant, K, for the following reaction is
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K.
2HI(g) -->
H2(g) +
I2(g)
An equilibrium mixture of the three gases in a 1.00 L flask at
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HI,
4.33×10-2 M
H2 and
4.33×10-2 M
I2. What will be the concentrations of
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[HI]
=
M
[H2]
=
M
[I2]
=
M

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2.27×10-2 mol of I2(g) is added to the flask?
[HI] = _____M
[H2] = ____M
[I2] = _____M

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