NH4HS(s)NH3(g)+H2S(g) An empty 5.00-L flask is charged with 0.450 g of pure H2S(

Question

NH4HS(s)NH3(g)+H2S(g)

An empty 5.00-L flask is charged with 0.450 g of pure H2S(g), at 25 C.

Part A

Part complete

Evaluate the validity of the following statements.

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True

No reaction occurs.

The pressure of ammonia is zero.

Q is equal to 0.

False

Q is greater than K.

The reaction proceeds to the left.

The reaction proceeds to the right.

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Correct

Part B

Part complete

What is the initial pressure of H2S(g) in the flask?

Express your answer numerically in atmospheres.

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Correct

Significant Figures Feedback: Your answer 0.0648 atm was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your answer.

Addition of ammonium bisulfate

In addition to the H2S already present in the flask, solid NH4HS is added until there is excess unreacted solid remaining.

Part C

What are the partial pressures of NH3 and H2S at equilibrium, that is, what are the values of PNH3 and PH2S, respectively?

Enter the partial pressure of ammonia followed by the partial pressure of hydrogen sulfide numerically in atmospheres separated by a comma.

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Part D

What is the mole fraction, , of H2S in the gas mixture at equilibrium?

Express your answer numerically.

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Part E

What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.450 g of pure H2S(g), at 25 C to achieve equilibrium?

Express your answer numerically in grams.

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Ammonium bisulfide, NH4HS, forms ammonia, NH3, and hydrogen sulfide, H2S, through the reaction

NH4HS(s)NH3(g)+H2S(g)

This reaction has a Kp value of 0.120 at 25 C.

An empty 5.00-L flask is charged with 0.450 g of pure H2S(g), at 25 C.

Part A

Part complete

Evaluate the validity of the following statements.

Drag each item to the appropriate bin.

View Available Hint(s)

Reset

Help

True

No reaction occurs.

The pressure of ammonia is zero.

Q is equal to 0.

False

Q is greater than K.

The reaction proceeds to the left.

The reaction proceeds to the right.

SubmitPrevious Answers

Correct

Part B

Part complete

What is the initial pressure of H2S(g) in the flask?

Express your answer numerically in atmospheres.

View Available Hint(s)

P = 6.46×102   atm  

SubmitPrevious Answers

Correct

Significant Figures Feedback: Your answer 0.0648 atm was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your answer.

Addition of ammonium bisulfate

In addition to the H2S already present in the flask, solid NH4HS is added until there is excess unreacted solid remaining.

Part C

What are the partial pressures of NH3 and H2S at equilibrium, that is, what are the values of PNH3 and PH2S, respectively?

Enter the partial pressure of ammonia followed by the partial pressure of hydrogen sulfide numerically in atmospheres separated by a comma.

View Available Hint(s)

PNH3, PH2S =   atm  

Submit

Part D

What is the mole fraction, , of H2S in the gas mixture at equilibrium?

Express your answer numerically.

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H2S =

Submit

Part E

What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.450 g of pure H2S(g), at 25 C to achieve equilibrium?

Express your answer numerically in grams.

View Available Hint(s)

mass>   g  

Submit

Explanation / Answer

Part A:

Since only H2S is present in reaction mixture initially no reaction will occur.

Q is equal to 0 TRUE

Q is greater than K FALSE because Q = K = 0

No reaction occurs TRUE

The reaction proceeds to the left FALSE because no reaction occurs

The reaction proceeds to the right FALSE

The pressure of ammonia is zero TRUE because no ammonia is present initially or formed

Part B)

Let us consider Ideal gas equation

PV = nRT

pressure, P = nRT/V

Volume , V = 5L

Temperature,T = 25 = 298.15K

No of mole , n = 0.450g/34.079g/mol = 0.01320

gas constant ,R = 0.082057(L atm /mol K )

Therefore,

P = 0.01320 mol × 0.082057(L atm /mol K ) × 298.15K/5L

= 0.0646atm

Part C

Temperature = 25 oC = 298 K

Volume = 5 L

Mass of H2S = 0.450 g

Moles of H2S = 0.450 / 34.08

=0.0132

Using PV = nRT

Initial pressure of H2S = 0.0132 * 0.0821 * 298 / 5

=0.0645 atm

Kp = 0.120

NH4HS(s)           NH3(g)      +       H2S(g)
Initial                             -                         0.0                     0.0645
Change                         -                          +x                       +x
Equilibrium                   -                           x                    0.0645 + x

Kp = P(NH3) * P(H2S)

0.120 = x (0.0645 + x)

x = 0.315 atm

Hence, at equilibrium:

Partial pressure of NH3 = 0.315 atm

Partial pressure of H2S = 0.0645 + 0.315

= 0.3795 atm

Part D)

Total pressure at equilibrium = 0.315 + 0.3795

= 0.694 atm

We know that:

Partial pressure = mole fraction * total pressure

0.3795 = X * 0.694

X = 0.3795 / 0.694

X = 0.546

Part E)

Moles of H2S at equilibrium = PV /RT

= 0.3795 * 5 / 0.0821 * 298

= 0.077

Moles of H2S produced from NH4HS = 0.077 - 0.0132

= 0.0638

Moles of NH4HS required = 0.0638

Minimum mass of NH4HS that must be added = 0.0638 * 51.11

= 3.26 g

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