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

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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.

Chemistry

2 answers:

Juli2301 [7.4K] · Answer 1 · 2022-02-25T17:51:05+03:00

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

kari74 [83] · Answer 2 · 2022-02-25T19:29:30+03:00

Answer:

[HI] = 0.83 M

[H₂] = 0.070 M

[I₂] = 0.18 M

Explanation:

We have the equilibrium:

H₂ (g) + I₂ (g) ⇄ 2 HI (g)

for which

Kc = 54.3 = [HI]² / ( [H₂][I₂]

The concentrations of the gases initially are:

[H₂]₀ = 0.714 mol / 2.40 L = 0.2975 M

[I₂]₀ = 0.984 mol / 2.40 L = 0.4100 M

[HI]₀ = 0.886 mol / 2.40 L = 0.3692 M

We should use Q since the question states we are not at equilibrium:

Q = [HI]² / ( [H₂][I₂] = 0.369²/ ( 0.298 x 0.410 ) = 1.15

Q is less than Kc ( 1.15 vs 54.3 ), so the system will shift to the right to attain equilibrium, that is the concentrations of H₂ and I₂ will decrease to favor HI

Lets call x the concentration decrease of H₂ and I₂, we can then write the equation for the equilibrium:

Kc = 54.3 = ( 0.369 + 2x )² / ( 0.298 - x ) ( 0.410 - x )

A quadratic equation results which we can solve from our calculator or making use of software:

<u>0.369² + 2 ( 0.369 x 2x ) + (2x)²</u> =54.3

x² - 0.708 x + 0.122

0.136 + 1.476x + 4x² = 54.3 ( x² - 0.708 x + 0.122 )

0.136 + 1.476x + 4x² = 54.3 x² - 38.444 + 6.625

0 = 50.3 x² - 39.920 x +6.489

The roots are X₁ = 0.566 and X₂ = 0.228

The first is impossible since it will give negative concentrations at equilibrium for H₂ and I₂.

So using X₂ = 0.228

[HI] = 0.369 + 2(0.228) = 0.83 M