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Serggg [28]
3 years ago
15

The rate constant for the first-order decomposition of N2O5 (g) to NO2 (g) and O2 (g) is 7.48 * 10-3 s-1 at a given temperature.

Chemistry
1 answer:
alina1380 [7]3 years ago
8 0

Answer:

Explanation:

In this problem, we have a first-order decomposition reaction with a given rate constant. The rate law for a first-order reaction like this is

r

a

t

e

=

k

[

A

]

, where k is the rate constant and [A] is the concentration of the reactant (renamed as A, for brevity). To find the dynamics of the reaction with time, we can integrate the rate law to get an expression for [A](t):

rate = −d[A]dt = k[A]

[A]f =[A]i e−kt

We want the total pressure of the reaction chamber to be 0.145 atm, with a starting reactant pressure of 0.110 atm. To solve for the time this reaction takes, we need the reaction equation:

2N2O5(g) → 4NO2(g) + O2(g)

Using the stoichiometry of the reaction equation, we can determine the final pressure of the reactant. This requires us to rewrite the total pressure equation in terms of the change in pressure of the reactant.

Pf=0.145atm

Pi=0.110atm = pN2O5

iPf =pN2O5

f +pNO2 + pO2pNO2 = 4pO2

This comes from the stoichiometry.

pNO2 = 2(pN2O5i − pN2O5f )

This comes from the stoichiometry.

pNO2 = −2ΔpN2O5Pf = (pN2O5i + ΔpN2O5) − 2ΔpN2O 5 − 12ΔpN2O5

0.145atm =(0.110atm + ΔpN2O5) − 2.5ΔpN2O5 = 0.110atm − 1.5ΔpN2O5

ΔpN2O5 = −0.0233atm

pN2O5f = 0.110atm − 0.0233atm =

0.0867atm

This is our final pressure! Now we can use the integrated rate law.

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What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at
Alchen [17]

Answer:

4,313.43 mmHg is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at  850 mm Hg

Explanation:

Boyle's law says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." This means that if the quantity of gas and the temperature remain constant, the product of the pressure for the volume always has the same value.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

If you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment and you vary the volume of gas to a new value V2, then the pressure will change to P2, it will be true:

P1 * V1 = P2 * V2

In this case:

  • V1=0.335 L
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Replacing:

P1*0.335 L=850 mmHg*1.7 L

Solving:

P1=\frac{850 mmHg*1.7 L}{0.335 L}

<u><em>P1=4,313.43 mmHg</em></u>

<u><em>4,313.43 mmHg is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at  850 mm Hg</em></u>

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Answer:

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Explanation:

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