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50.0g of N2O4 is introduced into an evacuated 2.00L vessel and allowed to come to equilibrium with its decomposition product,N2O

Lady_Fox [76]

The mass of N2O4 in the final equilibrium mixture is 39.45 grams.

The decomposition reaction of N2O4 to 2NO2 can be expressed as:

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

From the parameters given:

The mass of N2O4 = 50.0 g
The molar mass of N2O4 = 92.011 g/mol

The number of mole of N2O4 can be determined as:

$\mathbf{Moles \ of \ N_2O_4 = \dfrac{50.0 g}{92.011 g/mol}}$

$\mathbf{Moles \ of \ N_2O_4 = 0.5434 \ moles}$

The volume of the vessel in which N2O4 was evacuated is = 2.0 L

From stochiometry, the concentration of $\mathbf{[N2O4] = \dfrac{mole \ of \ N_2O_4}{volume \ of \ N_2O_4}}$

$\mathbf{[N2O4] = \dfrac{0.5434}{2}}$

$\mathbf{[N2O4] = 0.2717 \ M}$

The I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.2717 0

Change -x +2x

Equilibrium (0.2717 - x) 2x

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(2x)^2}{(0.2717-x)}}$

Recall that; Kc = 0.133

∴

$\mathbf{0.133 = \dfrac{4x^2}{(0.2717-x)}}$

0.0361 - 0.133x = 4x²

4x² + 0.133x - 0.0361 = 0

By solving the above quadratic equation, we have;

x = 0.07978

The Concentration of [NO2] = 2x

[NO2] = 2 (0.07978)
[NO2] = 0.15956 M

The Concentration of [N2O4] = 0.2717 - x

[N2O4] = 0.2717 - 0.07978
[NO2] = 0.19192 M

Again, from the decomposition reaction, we can assert that;

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

0.19192 M of N2O4 decompose to produce 0.15956 M of NO2.

However, if 5.0 g of NO2 is injected into the vessel, then the number of moles of NO2 injected becomes;

$\mathbf{= \dfrac{5.0 \ g}{46 \ g/mol}} \\ \\ \\ \mathbf{= 0.10869\ moles}$

The Molarity of NO2 injected now becomes:

$\mathbf{= \dfrac{00.10869 }{2} } \\ \\ \\ \mathbf{= 0.05434 \ M }$

So, the new moles of [NO2] becomes = 0.15956 + 0.05434

= 0.2139 M

The new I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.19192 0.2139 M

Change +x -2x

Equilibrium (0.19192 + x) (0.2139 -2x)

NOTE: The injection of NO2 makes the reaction proceed in the backward direction.

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

Recall that; Kc = 0.133

$\mathbf{0.133= \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

By solving for x;

x = 0.2246 or x = 0.0225

We need to consider the value of x that is less than the initial concentration of NO2(0.2139 M) which is:

x = 0.0225

Now, the final concentration of [N2O4] = (0.19192 + 0.0225)M

= 0.21442 M

The final number of moles of N2O4 = Molarity(concentration) × volume

The final number of moles of N2O4 = (0.21442 × 2) moles

The final number of moles of N2O4 = 0.42884 moles

The mass of N2O4 in the final equilibrium mixture is:

= final number of moles × molar mass of N2O4

= 0.42884 moles × 92 g/mol

= 39.45 grams

Learn more about the decomposition of N2O4 here:

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6 0

3 years ago

Which action(s) is/are true of antihistamines? (select all that apply.) test bank?

Sav [38]

The correct answers from the choices are:

a. Reduce inflammation locally
b. Antagonize H1 receptors
c. May be administered orally
d. Are systemically distributed

Antihistamines are drugs that are H1 antagonists or blockers, may be administered orally, are distributed systemically, and do not directly reduce nasal congestion.

7 0

3 years ago

The splitting of carbon dioxide to form oxygen gas and carbon compounds occurs during

Oxana [17]

Answer:

neither photosynthesis nor respiration.

Explanation:

Photosynthesis is the process by which the green plants, in the presence of sunlight makes their own food. In the process carbon dioxide is absorbed and oxygen is released.

$6CO_2 + 6H_2O\ (sunlight)\ \rightarrow C_6H_{12}O_6 + 6O_2$