Question

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

Answer #1

An equilibrium mixture contains 0.710 mol HI, 0.460 mol I2, and
0.250 mol H2 in a 1.00-L flask.
What is the equilibrium constant for the following reaction?
2HI(g) H2(g) + I2(g)
K =
How many moles of I2 must be
removed in order to double the number of moles of H2 at
equilibrium?
_______ mol I2

Hydrogen iodide undergoes decomposition according to the
equation 2HI(g) H2(g) + I2(g) The equilibrium constant Kp at 500 K
for this equilibrium is 0.060. Suppose 0.898 mol of HI is placed in
a 1.00-L container at 500 K. What is the equilibrium concentration
of H2(g)?
(R = 0.0821 L · atm/(K · mol))
A. 7.3 M
B. 0.40 M
C. 0.15 M
D. 0.18 M
E. 0.043 M

The equilibrium constant, K, for the following reaction is
1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of
the three gases in a 1.00 L flask at 698 K contains 0.325 M HI,
4.36×10-2 M H2 and 4.36×10-2 M I2. What will be the concentrations
of the three gases once equilibrium has been reestablished, if
2.27×10-2 mol of I2(g) is added to the flask?
[HI] = _____M
[H2] = ____M
[I2] = _____M

. If 2.00 mol of H2 and 1.00 mol of I2 come to equilibrium at
this temperature (458ºC), how many moles of HI will be in the final
mixture? (First, consider this question: Why is the volume of the
container not needed for this problem?)
H2(g) + I2(g) 2HI(g) Kc = 50.3 at 458ºC

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.

For the reaction 2HI(g) ⇌H2(g)+I2(g), Kc= 0.290 at 400 K. If
the initial concentration of HI is 4.0×10−3M and the initial
concentrations of H2, and the initial concentrations of H2, and
I2 are both 1.50×10–3M at 400 K, which one of the following
statements is correct?
a. The concentrations of HI and I2 will increase as the system
is approaching equilibrium.
b. The concentrations of H2 and I2 will increase as the system
is approaching equilibrium.
c. The system is...

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?

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...

A 1.00 L container holds 0.015 mol of H2 (g) , 0.015 mol of I2
(g), and 0.015 mol of HI (g) at 721 K. What are the
concentrations(pressures) of H2 (g), I2 (g), and HI (g) after the
system achieved a state of equilibrium? The value of Kc is 50.0 for
reaction: H2 (g) + I2 (g) 2HI (g)

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?

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