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Anastasy [175]
3 years ago
8

Be sure to answer all parts. Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO an

d 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
Chemistry
2 answers:
Ivenika [448]3 years ago
8 0

Answer:

The remain gases are o_{2}_{(g)} and NO_{2}_{(g)}

Pressure of O_{2}_{(g)} 1.09 atm O_{2}_{(g)}

Pressure of NO_{2}_{(g)} 1.09 atm NO_{2}_{(g)}

Explanation:

We have the following reaction

2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}

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

PV= nRT

We cleared the mols

n=\frac{PV}{RT}

PV=nrT

replace the data for each gas

Constant of ideal gases

R= 0.082\frac{atm.l}{mol.K}

Transform degrees celsius to kelvin

25+273=298K

NO_{g}

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

O_{2}_{(g)

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

Find the limit reagent by stoichiometry

2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}

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

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

NO_{(g)}

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)}

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

Calculate the moles that are formed of NO_{2}_{(g)}  

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)}

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

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

0.086-0.04= 0.046

The remain gases areO_{2}_{(g)} and NO_{2}_{(g)}

calculate the volume that gases occupy  

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

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

Calculate partial pressures with the ideal gas equation

PV= nRT

P=\frac{nRT}{V}

Pressure of O_{2}_{(g)}

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

Pressure of NO_{2}_{(g)}

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

Verdich [7]3 years ago
5 0

Answer:

It will remain O₂ and NO₂, with partial pressures: pO₂ = 0.186 atm, and pNO₂ = 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

O₂:

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 O₂

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

By a simple direct three rule:

2x = 0.0798

x = 0.0399 mol of O₂

The number of moles of oxygen needed is lower than the number of moles in the reaction, so O₂ is the limiting reactant, and NO will be totally consumed. The number of moles of NO₂ formed will be:

2 moles of NO --------------- 2 moles of NO₂

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

By a simple direct three rule:

x = 0.0798 mol of NO₂

And the number of moles of O₂ that remains is the initial less the total that reacts:

n = 0.0855 - 0.0399

n = 0.0456 mol of O₂

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

O₂:

P = (0.0456*0.082*298)/5.99

P = 0.186 atm

NO₂:

P = (0.0798*0.082*298)/5.99

P = 0.325 atm

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