Question

7. The equilibrium constant Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 54.3 at 430°C. At the start of the reaction there are 0.714 moles of hydrogen, 0.984 moles of iodine, and 0.886 moles of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium.

Answer 1

Answer:

[H2] = 0.0692 M

[I2] = 0.182 M

[HI] =  0.826 M

Explanation:

Step 1: Data given

Kc = 54.3 at 430 °C

Number of moles hydrogen = 0.714 moles

Number of moles iodine = 0.984 moles

Number of moles HI = 0.886 moles

Volume = 2.40 L

Step 2: The balanced equation

H2 + I2 → 2HI

Step 3: Calculate Q

If we know Q, we know in what direction the reaction will go

Q = [HI]² / [I2][H2]

Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Q =(n(HI)²) /(nH2 *nI2)

Q = 0.886²/(0.714*0.984)

Q =1.117

Q<Kc This means the reaction goes to the right (side of products)

Step 2: Calculate moles at equilibrium

For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI

Moles H2 = 0.714 - X

Moles I2 = 0.984 -X

Moles HI = 0.886 + 2X

Step 3: Define Kc

Kc = [HI]² / [I2][H2]

Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]

Kc =(n(HI)²) /(nH2 *nI2)

KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))

X = 0.548

Step 4: Calculate concentrations at the equilibrium

[H2] = (0.714-0.548) / 2.40 = 0.0692 M

[I2] = (0.984 - 0.548) / 2.40 = 0.182 M

[HI] = (0.886+2*0.548) /2.40 = 0.826 M

Answer 2

Following are the calculation of the concentrations of the gases:

Given:

moles of

moles of $I_2= 0.984 \ moles$

moles of $HI = 0.886 \ moles$

$V = 2.40\ L$

To find:

concentrations of the gas=?

Solution:

Equation:

$\to \bold{H_2 + I_2 \leftrightharpoons 2HI}$

Calculating the reaction quotient (Q):

$\to Q = \frac{[HI]^2}{[I_2][H_2]}$

        $= \frac{[\frac{n(HI)}{V}]^2}{[\frac{n(I_2)}{V}][\frac{n(H_2)}{V}]}\\\\ = \frac{n(HI)^2}{n(H_2) \times n(I_2)}\\\\= \frac{0.886^2}{(0.714\times 0.984)}\\\\= \frac{0.784996}{0.702576}\\\\=1.117$

Therefore, It signifies that reaction is directed to the right.  

Calculating the moles in equilibrium:  

As per the given equation:

Mole of

Mole of  

Mole of

 Calculating the Kc:

$\to Kc = \frac{[HI]^2}{[I_2][H_2]}$

          $= \frac{[\frac{n(HI)}{V}]^2}{[\frac{n(I_2)}{V}][\frac{n(H_2)}{V}]}\\\\ = \frac{n(HI)^2}{n(H_2) \times n(I_2)}$

$\to 54.3= \frac{0.886 +2X}{(0.714-X) \times (0.984- X)}\\\\\to X=0.548$

 Calculating  the concentrations of the equilibrium:

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