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gulaghasi [49]
3 years ago
15

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

emist studying this reaction fills a flask with of sulfur dioxide gas and of oxygen gas, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be . Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to significant digits.
Chemistry
1 answer:
Inessa [10]3 years ago
8 0

Answer:

The answer is "\bold{4.97 \times 10^{-2}}"

Explanation:

Please find the complete question in the attached file.

Equation:

2SO_2+O_2  \leftrightharpoons 2SO_3

at t=0 3.3   \ \ \ \ \ \ \ \ \ \ 0.79

at equilibrium 3.3-p \ \ \ \ \ \ \ \ \ \ 0.79 - \frac{P}{2} \ \ \ \ \ \ \ \ \ \ \ \ P

p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\

     =\frac{0.47^2}{2.83^2\times 0.555}\\\\=4.97 \times 10^{-2}

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When dinitrogen pentaoxide, a white solid, is heated, it decomposes to produce nitrogen dioxide gas and
AysviL [449]

9.5314 L is the volume of nitrogen dioxide formed at 103.25 kPa and 22.75 °C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given data:

Oxygen produced - 1.618 gram

Decomposition of N_2O_5 takes place.

Find - Amount of NO_2 produced.

The decomposition reaction is as follows -

2N_2O_5--> 4NO_2 + O_2

Moles of O_2 gas =\frac{1.6}{16}  =0.1 moles.

1 mole of O_2 is produced from 2 moles of dinitrogen pentoxide

0.1 mole of O_2  will be produced from = 0.2 moles.

Now, 2 moles of dinitrogen pentoxide produce 4 moles of NO_2

NO_2 produced will be - 0.4 moles.

Weight of NO_2 produced - 0.4 X 46

Weight of NO_2  produced - 18.4 gram

Thus, grams of NO_2 produced are 18.4

Now calculate the volume of NO_2

Given data are:

P=103.25 kPa =1.01899827 atm

T= 22.75 °C +273 = 295.75 K

n=0.4 moles

V=?

R= 0.0821 liter·atm/mol·K

Putting the value in PV=nRT

V =  \frac{nRT}{P}

V =  \frac{0.4 \;moles \;X \;0.0821\; liter\;atm/\;mol \;K X \;295.75 \;K}{1.01899827 atm}

V= 9.5314 L

Hence, 9.5314 L is the volume of nitrogen dioxide formed at 103.25 kPa and 22.75 °C.

Learn more about the ideal gas equation here:

brainly.com/question/13450124

#SPJ1

8 0
2 years ago
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