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Sav [38]
3 years ago
13

At a certain temperature, the Kp for the decomposition of H2S is 0.784. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H 2 S H2S is presen

t at a pressure of 0.292 atm 0.292 atm in a closed container. What is the total pressure in the container at equilibrium?
Chemistry
1 answer:
user100 [1]3 years ago
3 0

Answer:

P = 0.557 atm

Explanation:

Let's write the equation again:

H₂S(g) <---------> H₂(g) + S(g)     Kp = 0.784

We know that only H2S is present at the beggining of reaction, and we want to know the total pressure in the container at equilibrium.

To do this, we need to know the partial pressure of all gases in equilibrium. To do that, we use an ICE chart, and solve for the partial pressure in equilibrium. Once we have that, we just sum the value of the pressure in equilibrium

Doing an ICE chart we have:

          H₂S(g) <---------> H₂(g) + S(g)     Kp = 0.784

I:         0.292                    0       0

C:           -x                       +x      +x

E:       0.292-x                  x       x

Writting the expression for Kp:

Kp = [H₂] [S] / [H₂S] --> replacing the values of the chart:

0.784 = x² / 0.292-x     solving for x

0.784(0.292-x) = x²

0.2289 - 0.784x = x²

x² + 0.784x - 0.2289 = 0

Using the general formula for x, in a quadratic equation we have:

x = -b ±√b² - 4ac / 2a

From the equation, we replace the values of a, b and c, and solve for x:

a = 1; b = 0.784; c = -0.2289

x = -0.784 ±√(0.784)² - 4 * 1 * (-0.2289) / 2 * 1

x = -0.784 ±√0.6147 + 0.9156 / 2

x = -0.784 ±√1.5303 / 2

x = -0.784 ± 1.237 / 2

x1 = -0.784 + 1.237 / 2 = 0.2265

x2 = -0.784 - 1.237 / 2 = -1.0105

The negative value is neglected, so we use x1 = 0.2265

Therefore the equilibrium pressures are:

PpH₂S = 0.292 - 0.2265 = 0.0655 atm

PpH₂ = PpS = 0.2265 atm

So, the total pressure in the container would be:

P = 0.0655 + 0.2265 + 0.2265

P = 0.557 atm

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Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

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The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

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ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

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ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

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