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

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(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

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atm/(K · mol))
2HI(g) H2(g) + I2(g)
A. 0.045
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C. 22
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What is the equilibrium concentration of I2?
M
What is the equilibrium concentration of I?
M

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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
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