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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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For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of
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Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For Al_2O_3

Mass of Al_2O_3  = 4.07 g

Molar mass of Al_2O_3  = 101.96 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{4.07\ g}{101.96\ g/mol}

Moles\ of\ Al_2O_3= 0.0399\ mol

According to the reaction:

Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O

1 mole of Al_2O_3  on reaction produces 1 mole of Al_2(SO_4)_3

So,  

0.0399 mole of Al_2O_3  on reaction produces 0.0399 mole of Al_2(SO_4)_3

Moles of Al_2(SO_4)_3  obtained = 0.0399 mole

Molar mass of Al_2(SO_4)_3 = 342.2 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0399= \frac{Mass}{342.2\ g/mol}

Mass= 13.7\ g

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

\%\ yield =\frac{10.4}{13.7}\times 100

<u>% yield =76 %</u>

6 0
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