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3 years ago

Correct first answer= Brianliest!PLEASE WOULD REALLY APPRECIATE IF YOU’D HELP ME OUT:’(

Chemistry

1 answer:

nalin [4]3 years ago

4 0

Answer:

The partial pressure of carbon dioxide in the flask is 0,69 atm and the total pressure in the flask is 1,62 atm

Explanation:

Let think the Ideal gas equation

Pressure . volume = n.R.T

where R = 0,082 L.atm/ mol.K

T is temperature in K (T in °C + 273)

n = moles

In a mixture you have to think P as Total Pressure and n, the total moles (mol from any gas in the mixture)

You have mass, so let's get the moles

H2 = 0,545g/ 2,016 g/m = 0,270 moles

CO2 = 8,89 g/44 g/m = 0,202 moles

Total moles = 0,270 + 0,202 = 0,472 moles

Pressure = (0,472 moles . 0,082L.atm/mol.K . 347K) / 8,30L

Pressure = 1,62 atm

Pressure of CO2 must be calculated by the same rule, just take account moles from CO2

Pressure = (0,202 moles . 0,082L.atm/mol.K . 347K) / 8,30L

Pressure = 0,692 atm

There is another way to solve the partial pressure which involves the molar fraction.

Pressure in one gas / Total pressure = n moles in one gas / Total moles

Pressure CO2 / Total pressure = n moles CO2 / Total moles

Pressure CO2 = (n moles CO2/Total moles). Total pressure

Pressure CO2 = (0,202 moles / 0,472 moles). 1,62 atm

Pressure CO2 = 0,69 atm

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Explanation:

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Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

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Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

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$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

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