Question

Consider the reaction:

H_{2}(*g*)+I_{2}(*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*_{H}_{2}=*P*_{I2}=0.629
atm , and *P*_{HI}= 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 how to set up the equation to solve
for the partial pressure of HI for the second reaction.

Answer #1

We see that Qp > Kp for the reaction, hence the reaction will move in Backward direction

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

0.629....0.629.........0.101

+X.........+X.............-2X

0.629+X..0.629+X...0.101-2X

Kp = 4.76 x 10^-4 = (0.101-2X)^2 / (0.629+X)^2

Taking square roots on both sides

=> (0.101-2X) / (0.629+X) = 0.0218

=> 0.101 - 2X = 0.0137 + 0.0218 X

=> X = 0.0432 atm

**partial pressure of HI when the reaction reaches
equilibrium at 175 K = 0.101 - 2X = 0.0147 atm**

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 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 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.

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?
A.0.360 atm
B. 0.464 atm
C. 0.152 atm
D. 0.104...

The equation for the formation of hydrogen iodide from
H2 and I2 is:
H2(g) + I2(g) <--> 2HI(g)
The value of Kp for the reaction is 69.0 at 730.0C.
What is the equilibrium partial pressure of HI in a sealed reaction
vessel at 730.0C if the initial partial pressures of H2
and I2 are both 0.1600 atm and initially there is no HI
present?

2) Consider the reaction: I2 (s) + H2 (g) → 2 HI (g)
a) Calculate the value of the equilibrium constant Kp.
b) Calculate ΔGrxn at PHI = 2.55 atm, PH2 = 0.221 atm. Consider
excess I2(s)

At 400 K, an equilibrium mixture of H2, I2, and HI consists of
0.082 mol H2, 0.084 mol I2, and 0.15 mol HI in a 2.50-L flask. What
is the value of Kp for the following equilibrium? (R = 0.0821 L ·
atm/(K · mol))
2HI(g) H2(g) + I2(g)
A. 0.045
B. 7.0
C. 22
D. 0.29
E. 3.4

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