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finlep [7]
3 years ago
14

Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°

C and 0.94 atm, the density of this gas mixture is 2.7 g·L−1. What is the partial pressure of each gas? (2 sig fig)
Chemistry
1 answer:
Pavel [41]3 years ago
8 0

Answer:

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that  involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

PV = \frac{m}{M}RT

PxM = \frac{m}{V}RT

PxM = ρRT

ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

2.7 = \frac{0.94xMavg}{0.082x289}

0.94Mavg = 63.9846

Mavg = 68.0687 g/mol

The molar mass of N is 14 g/mol and of O is 16 g/mol, than M_{NO2} = 46 g/mol and M_{N2O4} = 96 g/mol. Calling y the molar fraction:

Mavg = M_{NO2}y_{NO2} + M_{N2O4}y_{N2O4}

And,

y_{NO2} + y_{N2O4} = 1

y_{N2O4} = 1 - y_{NO2}

So,

68.0687 = 46y_{NO2} + 92x(1 - y_{NO2})

68.0687 - 92 = 46y_{NO2} - 92y_{NO2}

46y_{NO2} = 23.9313

y_{NO2} = 0.52

y_{N2O4} = 0.48

The partial pressure is the molar fraction multiplied by the total pressure so:

PNO₂ = 0.52x0.94 = 0.49 atm

PN₂O₄ = 0.48x0.94 = 0.45 atm

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The given balanced equilibrium reaction is,

                  2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g)

Initial conc.          1 M                      0M          1 M

At eqm. conc.     (1-2x) M              (2x) M       (1+x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[NO]^2[Cl_2]}{[NOCl]^2}

The K_c= 1.6\times 10^{-5}}

Now put all the given values in this expression, we get :

{1.6\times 10^{-5}}=\frac{(2x)^2\times (1+x)}{(1-2x)^2}

By solving the term 'x', we get :

x=0.0019

Concentration of NO at equilibrium= (2x) M  =  2\times 0.0019=3.8\times 10^{-3}M

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For the chemical equation SO 2 ( g ) + NO 2 ( g ) − ⇀ ↽ − SO 3 ( g ) + NO ( g ) the equilibrium constant at a certain temperatur
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Answer : The number of moles of NO_2 gas added must be, 1.59 moles.

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

The given equilibrium reaction is,

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Initial conc.    2.86               x             0          0

At eqm.       (2.86-1.30)    (x-1.30)     1.30     1.30

The expression of K_{eq} will be,

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Now put all the given values in this expression, we get:

3.70=\frac{(1.30)\times (1.30)}{(2.86-1.30)\times (x-1.30)}

x=1.59

Thus, the number of moles of NO_2 gas added must be, 1.59 moles.

4 0
3 years ago
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