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Ozone reacts completely with NO, producing NO2 and O2. A 13.0 L vessel is filled with 1.30 mol of NO and 1.30 mol of O3 at 401.0 K. Find the partial pressure of each product and the total pressure in the flask at the end of the reaction.

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arsilan324

Answer:

6.58 atm total

3.29 atm NO2

3.29 atm O2

Explanation:

Balanced equation:

O3 + NO → NO2 + O2

There are equal numbers of moles of both reactants, so neither is in excess and either could be considered the limiting reactant.

( 1.30 mol NO) x (1 mol NO2 / 1 mol NO) =  1.30 mol NO2

( 1.30 mol NO) x (1 mol O2 / 1 mol NO) =  1.30 mol O2

Total pressure by using the formula;

P = nRT / V

= ( 1.30 mol +  1.30 mol) x (0.08205746 L atm/K mol) x (401.0 K) / (13.0 L)  

= 6.58 atm

Partial pressure for NO2;

(6.58 atm) x (1.30 mol NO2) / (1.30 mol + 1.30 mol)

= 3.29 atm NO2

Partial pressure for O2

6.58 atm total - 3.29 atm NO2

= 3.29 atm O2

Lanuel

1. The partial pressure of nitrogen dioxide ([tex]NO_2[/tex]) is equal to 3.29 atm.

2. The partial pressure of oxygen gas ([tex]O_2[/tex]) is equal to 3.29 atm.

3. The total pressure in the flask at the end of the chemical reaction is 6.58 atm.

Given the following data:

  • Volume of flask = 13.0 Liters
  • Number of moles of NO = 1.30 moles.
  • Number of moles of [tex]O_3[/tex] = 1.30 moles.
  • Temperature = 401.0 K.

Scientific data:

  • Ideal gas constant, R = 0.0821L⋅atm/mol⋅K

To determine the partial pressure of each product and the total pressure in the flask at the end of the chemical reaction:

First of all, we would write a balanced chemical equation for the chemical reaction as follows:

                                  [tex]O_3 + NO \rightarrow NO_2+O_2[/tex]

Since the numbers of moles of reactants are equal, the total number of moles of products is:

[tex]n=1.3+1.3[/tex]

n = 2.6 moles

Now, we can find the total pressure in the flask at the end of the chemical reaction by using the ideal gas law equation;

[tex]PV=n RT[/tex]

Where;

  • P is the pressure.
  • V is the volume.
  • n is the numbers of moles of gas.
  • R is the ideal gas constant.
  • T is the temperature.

Making P the subject of formula, we have;

[tex]P=\frac{nRT}{V}[/tex]

Substituting the parameters into the formula, we have;

[tex]P=\frac{2.6 \times 0.0821 \times 401}{13}\\\\P=\frac{85.597}{13}[/tex]

Total pressure, P = 6.58 atm.

Next, we would determine the partial pressure of each product:

[tex]Molefraction \;of \;a \;substance =\frac{No.\; of \; moles \;of \;substance}{Total \;no. \;of \; moles \;of \;substances}[/tex]

[tex]Molefraction \;of \;a \;substance =\frac{1.3}{2.6} \\\\Molefraction \;of \;a \;substance =0.5[/tex]

For [tex]NO_2[/tex]:

[tex]Partial \;pressure = Molefraction \times Total\;pressure\\\\Partial \;pressure = 0.5 \times 6.58[/tex]

Partial pressure of [tex]NO_2[/tex] = 3.29 atm.

For [tex]O_2[/tex]:

[tex]Partial \;pressure = Molefraction \times Total\;pressure\\\\Partial \;pressure = 0.5 \times 6.58[/tex]

Partial pressure of [tex]O_2[/tex] = 3.29 atm.

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