Chemistry, 22.02.2020 02:28 kelli151

If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2(g) 2NO(g) + O2(g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2.

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Answer from: Coltonh7681

The partial pressure of $\rm NO_2$ at equilibrium has been 0.152 atm.

The partial pressure has been the pressure that has been exerted by the single gas in the mixture. The given:

$\rm NO_2$ pressure = 0.5 atm.

The total pressure = 0.674 atm.

At the equilibrium, the reaction pressure has been given as:

$\rm 2\;NO_2\;\rightarrow\;2\;NO\;+\;O_2$

0.5 0 0 I

-2x 2x x C

0.5 - 2x +2x x E

According to the law of conservation:

Total pressure = sum of the partial pressure of each gas:

0.674 atm = 0.5 - 2x + 2x + x

0.674 atm = 0.5 + x

x = 0.174 atm.

The partial pressure of $\rm NO_2$ at equilibrium has been 0.5 - 2x

= 0.5 - 2 × 0.174 atm.

= 0.152 atm

The partial pressure of $\rm NO_2$ at equilibrium has been 0.152 atm.

For more information about the partial pressure, refer to the link:

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Answer from: kcceff4085

The partial pressure of NO2 = 0.152 atm

Explanation:

Step 1: Data given

Pressure NO2 = 0.500 atm

Total pressure at equilibrium = 0.674 atm

Step 2: The balanced equation

2NO2(g) → 2NO(g) + O2(g)

Step 3: The initial pressure

pNO2 = 0.500 atm

pNO = 0 atm

p O2 = 0 atm

Step 4: Calculate pressure at the equilibrium

For 2 moles NO2 we'll have 2 moles NO and 1 mol O2

pNO2 = 0.500 - 2x atm

pNO =2x atm

pO2 = xatm

The total pressure = p(total) = p(NO2) + p(NO) + p(O2)

p(total) = (0.500 - 2x) + 2x + x= 0.674 atm

0.500 + x = 0.674 atm

x = 0.174 atm

This means the partial pressure of NO2 = 0.500 - 2*0.174 = 0.152 atm

Answer from: Quest

um i think it might be a car?

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction ve...

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