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AleksandrR [38]
3 years ago
12

A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+

O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .
Chemistry
1 answer:
natima [27]3 years ago
3 0

Answer:

A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .

(4x^2)x

Kc= -----------

(A-2x)^2

PV=nRT

n/v = P/RT = .745/(0.0821)(334+273) = .01495

To Find the initial molarity of NO2

(mol/L)(g/mol) + (mol/L)(g/mol) + (mol/L)(g/mol)= g/L

Thus:

46(A-2x) + 2x(30) + 32x = .515 g/L

46A-92x+60x+32x = .515

46A=.515

A=.01120 M

Using the total molarity found

(A-2x)+2x+x = .01495 M

A+x=.01495

Plug in A found into the above equation:

.01120+x = .01495

x=.00375

Now Plug A and x into the original Equilibrium Constant Expression:

(4x^2)x

Kc= -----------

(A-2x)^2

Kc = 0.000014

Explanation:

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Enter your answer in the provided box.
nexus9112 [7]

Answer:

V₂ = 50.93 L

Explanation:

Initial volume, V_1=43.1\ L

Initial temperature, T_1=24^{\circ} C=24+273=297\ K

Final temperature, T_2=78^{\circ} C=78+273=351\ K

We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L

So, the final volume of the gas is 50.93 L.

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