Question

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?

Answer 1

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 .

Answer 2

273.0322 is the equilibrium constant, for the reaction.

The equilibrium constant is proportionate of the equilibrium concentration of the products to that of the reactants raised by the powers of the coefficients stoichiometry.

How to calculate the equilibrium concentration?

For the reaction, the ICE data is attached in the image below.

Given,

• The equilibrium concentration of the $\rm I_{2}$ = 0.0500 M
• The initial concentration of the $\rm I_{2}$ = 2.35 M
• The initial concentration of the $\rm H_{2}$ = 3.85 M

Step 1: Find the value of x

The concentration of  $\rm I_{2}$ ( ICE data) = Concentration of $\rm I_{2}$ (from the question )

$\begin{aligned}2.35 - \rm x &= 0.0500\\\\\rm x &= 2.3\end{aligned}$

Step 2: Calculate concentrations at equilibrium

For Hydrogen:

$\begin{aligned}\rm [H_{2}] &= 3.85 - \rm x\\\\\rm [H_{2}] &= 3.85 - 2.3\\\\\rm [H_{2}] &= 1.55 \;\rm M\end{aligned}$

For hydrogen iodide:

$\begin{aligned}\rm [HI] &= 2\rm x\\\\\rm [HI] &= 2 \times 2.3\\\\\rm [HI] &= 4.6 \;\rm M\end{aligned}$

Step 3: Calculate equilibrium constant

$\begin{aligned}\rm Kc &= \dfrac{[\rm HI]^{2}} { [\rm H_{2}][I_{2}]}\\\\\rm Kc &= \dfrac{( 4.6 )^{2} }{(1.55)\times(0.0500)}\\\\\rm Kc &= 273.0322 \end{aligned}$

Therefore, 273.0322 is the equilibrium constant, for the reaction.

Learn more about the equilibrium constant here:

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