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monitta
4 years ago
15

Nitrogen gas (N₂) reacts with hydrogen gas (H₂) to form ammonia (NH₃). At 200°C in a closed container, 1.0 atm of nitrogen gas i

s mixed with 2.0 atm of hydrogen gas. At equilibrium, the total pressure is 1.9 atm. Calculate the partial pressure of hydrogen gas at equilibrium.
Chemistry
1 answer:
Pie4 years ago
8 0

Answer:

The partial pressure of hydrogen gas at equilibrium is 1.26 atm

Explanation:

Let's use the molar fraction to solve this:

Molar fraction = Moles of gas / Total moles

Molar fraction = Gas pressure / Total pressure

Without equilibrium, we can think that the total system pressure is the sum of the partial pressures of each gas.

1 atm N₂ + 2 atm H₂ = 3 atm

Molar fraction for H₂ = 2 atm / 3atm → 0.66

Let's replace the molar fraction in equilibrium

Gas pressure / 1.9 atm = 0.66

Gas pressure = 1.26atm

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a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses
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Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

PV=nRT

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

P=n\\or\\\frac{P}{n}=constant

\frac{P_1}{n_1}=\frac{P_2}{n_2}

  • Initial conditions:

P_1=93kPa\\n_1=5\text{ puffs}

  • Final conditions: When additional 3 puffs of air is added

P_2=?kPa\\n_2=8\text{ puffs}

Putting the values, in above equation, we get

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3 years ago
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5 0
3 years ago
Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)
Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

S°(H₂O) = 69.91 J/K.mol

S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0
3 years ago
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