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devlian [24]
2 years ago
5

At night, an equilibrium reaction between two different nitrogen compounds generates N2O5N2O5 in the atmosphere, as represented

below.
Reaction 1: NO3(g)+NO2(g)⇄N2O5(g)   K=2.6×10−11NO3(g)+NO2(g)⇄N2O5(g)   K=2.6×10−11
During the day, solar radiation is absorbed by NO3(g), resulting in its decomposition. Which of the following best explains whether the equilibrium concentration of N2O5(g) in the atmosphere in the daytime is different from that in the nighttime, and why?
a) [N2O5] will be higher during the day, because the decomposition of NO3(g) results in an increase in the rate of production of N2O5(g).
b) [N2O5] will be higher during the day, because NO2(g) will be in excess, leading to an increase in the rate of production of N2O5(g)
c) [N2O5] will be higher at night, because the decomposition of NO3(g) in the daytime will result in an increase in the rate of consumption of N2O5(g) to reform NO3(g)
d) [N2O5] will be about the same at nighttime and daytime, because the amount of NO2(g) will not be changed and the equilibrium will not be affected.
Chemistry
1 answer:
kakasveta [241]2 years ago
3 0

The breakdown of NO3 in the day leads to more consumption of N2O5 in the day hence its concentration is greater at night than in the day.

<h3>What is equilibrium constant?</h3>

The equilibrium constant is a number that shows how much reactants are converted to products in a reaction. It is often shown as capital letter K in English.

Given the situation described in the question, we know that the concentration of N205 will be higher at night, because the decomposition of NO3(g) in the daytime will result in an increase in the rate of consumption of N2O5(g) to reform NO3(g).

Learn more about equilibrium constant: brainly.com/question/17960050

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4 0
3 years ago
What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

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

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

T_1= initial temperature = 26.5^oC=273+26.5=299.5K

T_2 = final temperature =-5.1^oC=273+(-5.1)=267.9K

Now put all the given values in this formula, we get

\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]

\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

P_1=26.5mmHg

Thus the vapor pressure of CS_2CS_2 in mmHg at 26.5 ∘C is 26.5

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4 0
3 years ago
Read 2 more answers
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6 0
3 years ago
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IgorLugansk [536]

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Q = 1252.82kJ

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Q = 1252.82kJ

The heat energy required to heat 179g of acetic acid from 25°C to 82.70°C is 1252.82kJ

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