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Be sure to answer all parts. Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO and O2 are separated in two different chambers connected by a valve. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end and calculate their partial pressures. Assume that the temperature remains constant at 25°C. Initial conditions are as follows: NO: 3.90 L, 0.500 atm O2: 2.09 L, 1.00 atm

Respuesta :

Answer:

The remain gases are [tex]o_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]

Pressure of [tex]O_{2}_{(g)}[/tex] [tex]1.09 atm O_{2}_{(g)}[/tex]

Pressure of [tex]NO_{2}_{(g)}[/tex] [tex]1.09 atm NO_{2}_{(g)}[/tex]

Explanation:

We have the following reaction

[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]

Now we calculate the limit reagents, to know which of the two gases is completely depleted and which one is in excess.

Excess gas will remain in the tank when the reagent limits have run out and the reaction ends.

To calculate the limit reagent, we must calculate the mols of each substance. We use the ideal gas equation

[tex]PV= nRT[/tex]

We cleared the mols

[tex]n=\frac{PV}{RT}[/tex]

PV=nrT

replace the data for each gas

Constant of ideal gases

[tex]R= 0.082\frac{atm.l}{mol.K}[/tex]

Transform degrees celsius to kelvin

[tex]25+273=298K[/tex]

[tex]NO_{g}[/tex]

[tex]n=\frac{0.500atm.3.90l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.080mol NO_{(g)} \\[/tex]

[tex]O_{2}_{(g)[/tex]

[tex]n=\frac{1atm.2.09l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.086mol O_{2}_{(g)} \\[/tex]

Find the limit reagent by stoichiometry

[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]

[tex]0.086mol O_{2}_(g).\frac{2mol NO_{(g)} }{1mol O_{2}_{(g)} } =0.17mol NO_{(g)}[/tex]

Using [tex]O_{2}_{(g)}[/tex]as the limit reagent produces more [tex]NO_{(g)}[/tex] than I have, so oxygen is my excess reagent and will remain when the reaction is over.

[tex]NO_{(g)}[/tex]

[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{1mol O_{2}_ {(g)} }{2mol NO_{(g)} } =0.04mol O_{2}_{(g)}[/tex]

Using [tex]NO_{(g)}[/tex] as the limit reagent produces less [tex]O_{2}_{(g)}[/tex] than I have, so [tex]NO_{(g)}[/tex]  is my excess reagent and will remain when the reaction is over.

Calculate the moles that are formed of [tex]NO_{2}_{(g)}[/tex]  

[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{2mol NO_{2}_ {(g)} }{2mol NO_{(g)} } =0.080mol NO_{2}_{(g)}[/tex]

We know that for all [tex]NO_{(g)}[/tex] to react, 0.04 mol [tex]O_{2}_{(g)}[/tex] is consumed.

we subtract the initial amount of [tex]O_{2}_{(g)}[/tex] less than necessary to complete the reaction. And that gives us the amount of mols that do not react.

[tex]0.086-0.04= 0.046[/tex]

The remain gases are[tex]O_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]

calculate the volume that gases occupy  

[tex]0.080 mol NO_{2}_{(g)} .\frac{22.4l NO_{2}_{(g)} }{1molNO_{2}_{(g)} }}  =1.79 lNO_{2}_{(g)}[/tex]

[tex]0.046 mol O_{2}_{(g)} .\frac{22.4 l O_{2}_{(g)} }{1molO_{2}_{(g)} }}  =1.03 l O_{2}_{(g)}[/tex]

Calculate partial pressures with the ideal gas equation

[tex]PV= nRT[/tex]

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

Pressure of [tex]O_{2}_{(g)}[/tex]

[tex]P=\frac{0.046mol.0.082\frac{atm.l}{K.mol} 298K}{1.03l}= 1.09 atmO_{2}_{(g)}[/tex]

Pressure of [tex]NO_{2}_{(g)}[/tex]

[tex]P=\frac{0.080mol.0.082\frac{atm.l}{K.mol} 298K}{1.79l}= 1.09 atmNO_{2}_{(g)}[/tex]

IthaloAbreu

Answer:

It will remain Oâ‚‚ and NOâ‚‚, with partial pressures: pOâ‚‚ = 0.186 atm, and pNOâ‚‚ = 0.325 atm.

Explanation:

First, let's identify the initial amount of each reactant using the ideal gas law:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas equation (0.082 atm.L/mol.K), and T is the temperature (25°C + 273 = 298 K).

n = PV/RT

NO:

n = (0.500*3.90)/(0.082*298)

n = 0.0798 mol

Oâ‚‚:

n = (1.00*2.09)/(0.082*298)

n = 0.0855 mol

By the stoichiometry of the reaction, we must found which reactant is limiting and which is in excess. The limiting reactant will be totally consumed. Thus, let's suppose that NO is the limiting reactant:

2 moles of NO ------------------ 1 mol of Oâ‚‚

0.0798 mol ------------------- x

By a simple direct three rule:

2x = 0.0798

x = 0.0399 mol of Oâ‚‚

The number of moles of oxygen needed is lower than the number of moles in the reaction, so Oâ‚‚ is the limiting reactant, and NO will be totally consumed. The number of moles of NOâ‚‚ formed will be:

2 moles of NO --------------- 2 moles of NOâ‚‚

0.0798 mol ---------------- x

By a simple direct three rule:

x = 0.0798 mol of NOâ‚‚

And the number of moles of Oâ‚‚ that remains is the initial less the total that reacts:

n = 0.0855 - 0.0399

n = 0.0456 mol of Oâ‚‚

The final volume will be the total volume of the containers, V = 3.90 + 2.09 = 5.99 L, so by the ideal gas law:

PV = nRT

P = nRT/V

Oâ‚‚:

P = (0.0456*0.082*298)/5.99

P = 0.186 atm

NOâ‚‚:

P = (0.0798*0.082*298)/5.99

P = 0.325 atm

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