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Lady bird [3.3K]
3 years ago
11

The emission of NO₂ by fossil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture. The urea r

educes NO (which oxidizes in air to form NO₂) according to the following reaction: 2CO(NH₂)₂(g)+4NO(g)+O₂(g)→4N₂(g)+2CO₂(g)+4H₂O(g) Suppose that the exhaust stream of an automobile has a flow rate of 2.37 L/s at 657 K and contains a partial pressure of NO of 14.0 torr .What total mass of urea is necessary to react completely with the NO formed during 8.1 hours of driving?Express answer in g.
Chemistry
1 answer:
liq [111]3 years ago
4 0

Answer:

There is 709.1 grams of urea needed.

Explanation:

<u>Step 1:</u> The balanced equation:

2CO(NH2)2(g)+4NO(g)+O2(g)→4N2(g)+2CO2(g)+4H2O(g)

<u>Step 2:</u> Data given

The exhaust stream of an automobile has a flow rate of 2.37 L/s

This happens at a temperature of 657 Kelvin

The partial pressure of NO is 14.0 torr

<u>Step 3:</u> Calculating the volume during 8.1 hours

V = 2.37L/s ∙ 8.1hr ∙3600s/hr = 69109.2L = 69.1092m³

<u>Step 4</u>: Calculate  the partial pressure of nitric oxide:

p = 14 torr ∙ 101325Pa/760torr = 1866.5Pa

<u>Step 5:</u> Calculating number of moles of NO

⇒ We use the ideal gas law P*V=n*R*T

n(NO) = P*V/R*T

with P= The partial pressure of NO = 1866.5 Pa

with V =the volume of NO = 69.1092m³

with R = the gas constant = 8.314472 Pa*m³/mol*K

with T = 657 Kelvin

n(NO) = 1866.5Pa ∙ 69.1092m³ / (8.314472Pa*m³/mol*K ∙ 657K)

= 23.61 mol es

<u>Step 6:</u> Calculate moles of urea

Since there is consumed 2 moles of urea per 4 moles of nitric oxide.

This means for 24.432 moles of NO, there is consumed 23.61 /2 = 11.806 moles of urea.

<u>Step 7</u>: Calculate mass of urea

m(CO(NH₂)₂) = 11.806 moles ∙ 60.06g/moles = 709.07g

There is 709.1 grams of urea needed.

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The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k
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Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

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P_1=102\ mmHg

T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol

And R is the Universal Gas Constant.

R=0.008 314 kJ/Kmol

ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165

Taking inverse log both side we get,

\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg

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