Question

Hydrogen iodide, HI, decomposes at moderate temperature according to the equation 2HI (g) H2 (g) + I2 (g) When 4.00 mol HI was placed in a 5.00-L vessel at 458C, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc for the decomposition of HI at this temperature?

Answer #1

concentration of HI = 4.00 / 5.00 = 0.800 M

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

0.800 0 0

0.80 - 2x x x

At equilibrium :

moles of I2 = 0.442 mol

concentration of I2 = 0.442 / 5 = 0.0884 M

[H2] = x = 0.0884 M

[HI] = 0.623 M

Kc = [H2] [I2] / [HI]^2

= (0.0884)^2 / (0.6232)^2

Kc = 0.0201

**value of Kc = 0.0201**

Hydrogen iodide decomposes according to the equation 2HI
(g)<---> H2 (g) + I2 (g) for which K= .0156 at 400 degrees
celsius. If 0.550 mol of HI was injected into 2.00L reaction vessel
at 400 degrees celsuis. Calculate the concentration of H2 at
equilibrium?

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

Hydrogen iodide decomposes according to the following reaction.
2 HI(g) equilibrium reaction arrow H2(g) + I2(g) A sealed 1.5 L
container initially holds 0.00615 mol H2, 0.00445 mol I2, and
0.0163 mol HI at 703 K. When equilibrium is reached, the
equilibrium concentration of H2(g) is 0.00364 M. What are the
equilibrium concentrations of HI(g) and I2(g)?

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?

Kc for the reaction of hydrogen and
iodine to produce hydrogen iodide.
H2(g) + I2(g) ⇌
2HI(g)
is 54.3 at 430°C. Calculate the
equilibrium concentrations of H2,
I2, and HI at 430°C if
the initial concentrations are [H2] =
[I2] = 0 M,
and [HI] = 0.483 M.

Kc for the reaction of hydrogen and iodine to produce hydrogen
iodide. H2(g) + I2(g) ⇌ 2HI(g) is 54.3 at 430 ° C. Calculate the
equilibrium concentrations of H2, I2, and HI at 430 ° C if the
initial concentrations are [H2] = [I2] = 0 M, and [HI] = 0.445
M.

At a certain temperature, the equilibrium constant, Kc for this
reaction is 53.3. H2(g)+I2(g) = 2HI(g) At this temperature, 0.300
mol of H2 and 0.300 mol of I2 were placed in a 1.00 L container to
react. What concentration of HI is present at equilibrium? View
comments (1)

The decomposition of hydrogen iodide, 2HI(g)= H2(g)+I2(g), has a
rate constant of 9.51*10-9 L/mol*s at 500K and Ea of 176 kJ/mol. At
what temperature will the rate constant be 1.10*10-5 L/mol*s?

At a certain temperature, the equilibrium constant, Kc, for this
reaction is 53.3.
H2(g) + I2(g) <----> 2HI(g)
Kc=53.3
At this temperature, 0.400 mol of H2 and 0.400 mol of I2 were
placed in a 1.00-L container to react. What concentration of HI is
present at equilibrium?

H2(g)+I2(g)⇌2HI(g) A reaction mixture in a 3.71 L flask at a
certain temperature initially contains 0.760 g H2 and 96.8 g I2. At
equilibrium, the flask contains 90.5 g HI. Calculate the
equilibrium constant (Kc) for the reaction at this temperature.

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