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Whitepunk [10]
3 years ago
9

The following reaction was performed in a sealed vessel at 760 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at

concentrations of [H2]=3.85M and [I2]=2.35M . The equilibrium concentration of I2 is 0.0500 M . What is the equilibrium constant, Kc, for the reaction at this temperature?
Chemistry
1 answer:
zubka84 [21]3 years ago
6 0

Answer:

The equilibrium constant is 273.0322

Explanation:

For the given chemical reaction ,

ICE table can be written as -

                             H₂(g)    +      I₂(g)    ⇄     2HI(g)

initial moles         3.85             2.35                -

at equilibrium      3.85 - x       2.35 - x           2x

From question , at equilibrium the concentration of I₂ = 0.0500 M

The concentration of I₂ ( ICE table ) = concentration of I₂ (given in question )

2.35 - x = 0.0500

x = 2.3

Putting the value of x in ICE table , to obtain the concentration  terms as-

[H₂] = 3.85 - x

[H₂] = 3.85 - 2.3

[H₂] = 1.55 M

[HI] = 2x

[HI] = 2* 2.3

[HI] = 4.6 M

[I₂] = 0.0500M       (Given)

Equilibrium Constant ( Kc )

The value of equilibrium constant is written as the concentration of the products each raised to the power of their respective stoichiometry by the concentration of reactants each raised to the power of their respective stoichiometry.

Kc = [HI]² / [H₂][I₂]

Kc = ( 4.6 )² / (1.55)*(0.0500)

Kc = 273.0322 .

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Explanation:

It seems the question is incomplete. But an internet search shows me these values for the question:

" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."

Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>

First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:

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