Question

Consider the equilibrium I2(g) ⇌ 2I(g) at 860 K. Suppose 0.054 mol I2(g) is placed in a rigid 0.48-L container at 860 K. At equilibrium, I2(g) is 2.52% dissociated into its products. What is Kp at this temperature?

Answer #1

Let's write the equilibrium reaction once again:

I_{2} <-----> 2I

With this reaction, we know that the expression for K is:

Kc = [I]^{2} / [I_{2}]

and Kp = Kc (RT)^{}

So, first we calculate Kc and then Kp. For Kc:

[I2] = 0.054 / 0.48 = 0.1125 M

The concentration for I is the 2.52% of dissociation: 0.1125 * 0.0252 = 0.002835 M

I_{2} <--------------------> 2I

i. 0.1125 0

e. 0.1125 - 0.002835 (2* 0.002835)

Kc = (2*0.002835)^{2} / (0.1125 - 0.002835)

Kc = 3.21x10^{-5} / 0.109665

Kc = 2.93x10^{-4}

now for Kp, let's calculate first :

= 2-1 = 1

Kp = 2.93x10^{-4} * 0.0821 * 860

Kp = 0.02067

If you need more explanations, or have some doubts, tell me in a comment and I'll gladly help you out.

Hope this helps

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

At 800 K the equilibrium constant for I2(g)←−→2I(g) is
Kc=3.1×10−5. If an equilibrium mixture in a 10.0-L vessel contains
2.66×10−2 g of I(g), how many grams of I2 are in the mixture?

The dissociation of molecular iodine into iodine atoms is
represented as I2(g) ⇌ 2I(g) At 1000 K, the equilibrium constant Kc
for the reaction is 3.80 × 10−5. Suppose you start with 0.0454 mol
of I2 in a 2.28−L flask at 1000 K. What are the concentrations of
the gases at equilibrium?
What is the equilibrium concentration of I2?
M
What is the equilibrium concentration of I?
M

The dissociation of molecular iodine into iodine atoms is
represented as I2(g) ⇌ 2I(g) At 1000 K, the equilibrium constant Kc
for the reaction is 3.80 × 10−5. Suppose you start with 0.0454 mol
of I2 in a 2.35−L flask at 1000 K. What are the concentrations of
the gases at equilibrium?
What is the equilibrium concentration of
I2?
M
What is the equilibrium concentration of I?
M

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

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) <-------> 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?

For the equilibrium 2IBr(g)?I2(g)+Br2(g) Kp=8.5×10?3 at 150 ?C.
If 2.3×10?2 atm of IBr is placed in a 2.0-L container, what is the
partial pressure of IBr after equilibrium is reached?

At 414 K, Kp = 18 for the reaction H2(g) + I2(g) ⇌ 2HI(g).
Initially, 74.7 g of HI (MW = 127.912 g/mol) is injected into an
evacuated 5.66-L rigid cylinder at 414 K. What is the total
pressure inside the cylinder when the system comes to equilibrium?
R = 0.0820568 L·atm/mol·K. Report your answer to three significant
figures.

. If 2.00 mol of H2 and 1.00 mol of I2 come to equilibrium at
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mixture? (First, consider this question: Why is the volume of the
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