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Sedbober [7]
3 years ago
10

Nitrogen and hydrogen react to form ammonia, like this:N2(g)+3H2(g)→2NH3(g)Use this chemical equation to answer the questions be

low.Suppose 135, mmol of N₂ and 405, mmol of H₂ are added to an empty flask, How much N₂ will be in the flask at equilibrium? a. Noneb. Some, but less than 135, mmol.c. 135,mmold. More than 135, mmol.Suppose 235, mmol of NH₃ are added to an empty flask, How much N₂ will be in the flask at equilibrium? a. Noneb. Some, but less than 118, mmol.c. 118,mmold. More than 118, mmol.
Chemistry
1 answer:
user100 [1]3 years ago
4 0

Answer:

Option A is correct, there will be no N2 left in the flask

Explanation:

Step 1 : Data given

Number of moles of N2 = 135 mmol = 0.135 mol

Number of moles of H2 = 405 mmol = 0.405 mol

Step 2: The reaction

N2(g)+3H2(g)→2NH3(g)

Step 3:

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

Both will completely react. There is no limiting reactant.

There will be produce 0.270 moles NH3.

Option A is correct, there will be no N2 left in the flask

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From the stoichiometry of the problem, 2.01  × 10^5 g of zinc phosphate is produced.

<h3>What is volume?</h3>

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Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

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2NO_2(g)\rightleftharpoons N_2O_4(g)

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3.0 atm                 0

At equilibrium

(3.0-2p)                 p

Equilibrium partial pressure of NO_2=2.1atm=3.0-2p

p = 0.45 atm

The value of equilibrium constant wil be given by :

K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}=\frac{p}{(3.0-2p)^2}

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After addition of 1.5 atm of nitrogen dioxide gas equilibrium reestablishes it self :

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After adding 1.5 atm of NO_2:

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At second equilibrium:'

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The expression of equilibrium can be written as:

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= (0.45+P) atm = (0.45 + 0.37 )atm = 0.82 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

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