Question

Consider the following reaction:

H2(*g*)+I2(*g*)⇌2HI(*g*)

A reaction mixture at equilibrium at 175 K
contains*P*H2=0.958atm, *P*I2=0.877atm, and
*P*HI=0.020atm. A second reaction mixture, also at 175 K,
contains *P*H2=*P*I2= 0.628 atm , and *P*HI=
0.107 atm .

A) If not, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?

Answer #1

i) Calculation of K from first reaction mixture

H2(g) + I2(g) <------> 2HI(g)

K = (P_{HI})^{2}/(P_{H2} ×
P_{I2})

K = (0.020atm)^{2}/(0.958atm × 0.877atm ) = 4.76
×10^{-4}

ii) Calculation of partial pressure of HI when second reaction mixture reaches equillibrium

Reaction quotient , Q =
(0.107atm)^{2}/(0.628atm)^{2} = 0.02903atm

Q is not equal to K , therefore , reaction mixture is not at equillbrium

Initial partial pressure

P_{H2} = 0.628

P_{I2} = 0.628

P_{HI} = 0.107

Change in concentration

P_{H2} = - x

P_{I2} = - x

P_{HI} = + 2x

Equillibrium concentration

P_{H2} = 0.628 - x

P_{I2} = 0.628 - x

P_{HI} = 0.107 + 2x

so,

(0.107 + 2x)^{2}/(0.628 -x)^{2} = 4.76
×10^{-4}

0.107 + 2x / 0.628 - x = 0.02182

0.107 + 2x =_{I} 0.013703 - 0.02182x

2.02182x = - 0.093297

x = - 0.04615

Therefore , when equillibrium is established

P_{H2} = 0.628 - (-0.04615) = 0.6742atm

P_{I2} = 0.628 - (-0.04615) = 0.6742atm

P_{HI} = 0.107 + (2 × - 0.04615) = 0.0147atm

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?

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...

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

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?

1) The equilibrium constant, Kp, for the following
reaction is 0.497 at 500 K:
PCl5(g) <---
----->PCl3(g) +
Cl2(g) (reversible)
Calculate the equilibrium partial pressures of all species when
PCl5(g) is introduced into an evacuated
flask at a pressure of 1.52 atm at
500 K.
PPCl5
=
_____atm
PPCl3
=
______atm
PCl2
=
______atm
2) The equilibrium constant, Kp, for the following
reaction is 55.6 at 698 K:
H2(g) +
I2(g) <----- ---->
2HI(g) (reversible)
Calculate the equilibrium partial pressures...

Consider the reaction: H2(g)+I2(g)⇌2HI(g)
A reaction mixture in a 3.64 −L flask at 500 K initially
contains 0.376 g H2 and 17.97 g I2. At equilibrium, the flask
contains 17.76 g HI.
Part A Calculate the equilibrium constant at this
temperature.
I keep getting 13413.06 and its not right I'm running out of tries,
please help me.

Consider the following reaction: H2(g)+I2(g)⇌2HI(g) A reaction
mixture in a 3.75 L flask at a certain temperature initially
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Consider the following reaction: H2(g)+I2(g)⇌2HI(g) A reaction
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contains 0.767 g H2 and 97.0 g I2. At equilibrium, the flask
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significant figures.

The system
H2(g) + I2(g) ⇌ 2HI(g
) is at equilibrium at a fixed temperature with a partial
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HI?
A.0.360 atm
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The equilibrium constant, K, for the following reaction is
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K.
2HI(g) -->
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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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0.213 mol of HI(g) is added to
the flask?
[HI]
=
M
[H2]
=
M
[I2]
=
M

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