Question

. 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

Answer #1

The volume is not needed since there in no net change in the number of moles. 2 moles of reactants gives 2 moles of product.

You can assume Volume as V and solve. V will cancel out and you would not need its value

The reaction is,

Reaction: H2 + I2 ----> 2 HI

Initial: 2 .... 1...........0

Change: -X.... -X.........+2X

Final: 2-X.....1-X.........2X

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

=> 50.3 = (2X)^2 / (2-X) (1-X)

=> 50.3 = 4X^2 / (X^2 - 3X + 2)

=> 50.3 X^2 - 150.9 X + 100.6 = 4X^2

=> 46.3 X^2 - 150.9 X + 100.6 = 0

Solving the above quadratic we get,

X = 0.935 M

**Moles of HI = 2X = 2 x 0.935 = 1.87 moles**

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

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)

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?

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)

At a certain temperature, the equilibrium constant, Kc, for this
reaction is 53.3.
H2(g) + I2(g) <-------> 2 HI(g)
Kc=53.3
At this temperature, 0.600 mol of H2 and 0.600 mol of I2 were
placed in a 1.00-L container to react. What concentration of HI is
present at equilibrium?

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

A student ran the following reaction in the laboratory at 742 K:
H2(g) + I2(g) <<------->>>2HI(g) When she introduced
0.202 moles of H2(g) and 0.225 moles of I2(g) into a 1.00 liter
container, she found the equilibrium concentration of HI(g) to be
0.331 M. Calculate the equilibrium constant, Kc, she obtained for
this reaction. Kc =

-Consider the following reaction:
2HI(g)
H2(g) +
I2(g)
If 3.69 moles of HI(g),
0.570 moles of H2, and
0.558 moles of I2 are
at equilibrium in a 16.6 L container at
818 K, the value of the equilibrium constant,
Kc, is_____________
.Consider the following reaction:
2NH3(g)
N2(g) +
3H2(g)
If 1.31×10-3 moles of
NH3(g), 0.681 moles of
N2, and 0.495 moles of
H2 are at equilibrium in a
18.6 L container at 893 K, the
value of the equilibrium constant,...

at a particular temperature, K= 1.00 x 102 for the
reaction H2 (g) + I2 (g) <-> 2HI
(g)
In an experiment, 1.00 mole of H2, 1.00 mole of
I2, and 1.00 mole of HI are introduced into a 1.00-L
container. calculate the concentrations of all species when
equilibrium is reached. please do step by step

0.250 moles of H2 and 0.250 moles of I2 were placed in a 1.00 L
flask at 500˚. The equilibrium constant, Kc, for the reaction
H2(g)+I2(g)-> 2HI(g) is 54.3. Calculate the equilibrium
concentrations of all species.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 4 minutes ago

asked 11 minutes ago

asked 24 minutes ago

asked 39 minutes ago

asked 49 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago