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posledela
2 years ago
13

Nitric oxide (NO) from car exhaust is a primary air pollutant. Calculate the equilibrium constant for the reaction

Chemistry
1 answer:
viva [34]2 years ago
4 0

This problem is asking for the equilibrium constant at two different temperatures by describing the chemical equilibrium between gaseous nitrogen, oxygen and nitrogen monoxide at 25 °C and 1496 °C as the room temperature and the typical temperature inside the cylinders of a car's engine respectively:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Thus, the calculated equilibrium constants turned out to be 6.19x10⁻³¹ and 9.87x10⁻⁵ at the aforementioned temperatures, respectively, according to the following work:

There is a relationship between the Gibbs free energy, enthalpy and entropy of the reaction, which leads to the equilibrium constant as shown below:

\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=-RT ln(K)

Which means we can calculate the enthalpy and entropy of reaction and subsequently the Gibbs free energy and equilibrium constant. In such a way, we calculate these two as follows, according to the enthalpies of formation and standard entropies of N₂(g), O₂(g) and NO(g) since these are assumed constant along the temperature range:

\Delta _rH=2*90.25 kJ/mol - (0 kJ/mol+0 kJ/mol)=180.5kJ/mol\\\\\Delta _rS=2*(0.211 kJ/mol*K)-(0.192kJ/mol*K+0.205kJ/mol*K)=0.025kJ/mol*K

Then, we calculate the Gibbs free energy of reaction at both 25 °C and 1496 °C:

\Delta _rG_{25\°C}=180.5-(25+298.15)*0.025=172.42kJ/mol\\\\\Delta _rG_{1496\°C}=180.5-(1496+298.15)*0.025=135.65kJ/mol

And finally, the equilibrium constants derived from the general Gibbs equation and Gibbs free energies of reaction:

K=exp(-\frac{\Delta _rG}{RT} )\\\\K_{25\°C}=exp[-\frac{172420 J/mol}{(8.3145\frac{J}{mol*K})(298.15K)} ]=6.19x10^{-31}\\\\K_{1496\°C}=exp[-\frac{135650J/mol}{(8.3145\frac{J}{mol*K})(1769K)} ]=9.87x10^{-5}

Learn more:

  • (Gibbs free energy) brainly.com/question/15213613
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White raven [17]

Answer:

A) It's correctly written

B) 77%

C) 835 calories

Explanation:

A) From online sources, we have number of calories as follows;

Fats: 9 calories per gram

Protein; 4 calories per gram

Carbs; 4 calories per gram

Total calories for each;

Total fat = 3 × 9 = 27 calories

Total protein = 3 × 4 = 12 calories

Total carbs = 32 × 4 = 128 calories

(sugar and dietary Fibre are classified as carbohydrates and so total carbs takes care of their calories).

Thus, total number of calories per serving = 27 + 12 + 128 = 167 calories per serving which is same as what is given.

B) percent from carbohydrates per serving = total calories from carbs/total number of calories per serving × 100% = 128/167 × 100% ≈ 77%

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8 0
3 years ago
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8 0
3 years ago
Two reasons why emerging diseases are especially harmful to humans. (is science)
Aleonysh [2.5K]

Answer:

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Explanation:

7 0
2 years ago
If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution
Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

\epsilon = molar absorptivity of this solution =5.56\times 10^4 M^{-1} cm^{-1}

0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c

c=1.4163\times 10^{-5} M=14.16 \mu M

(1\mu M=10^{-6} M)

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) c=1.4163\times 10^{-5} mol/L

1 L of solution contains 1.4163\times 10^{-5} moles of red dye.

Mass of 1.4163\times 10^{-5} moles of red dye:

1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g

(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100

red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = c=1.4163\times 10^{-5} M

Concentration of red solution after dilution = c'

c=c'\times 5

1.4163\times 10^{-5} M=c'\times 5

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The final concentration of the diluted solution is 2.83\times 10^{-6} M

8 0
3 years ago
How many molecules are in 2.50 moles of carbon dioxide?
podryga [215]

Answer:

1.5055×10²⁴ molecules

Explanation:

From the question given above, the following data were obtained:

Number of mole CO₂ = 2.5 moles

Number of molecules CO₂ =?

The number of molecules present in 2.5 moles CO₂ can be obtained as:

From Avogadro's hypothesis,

1 mole of CO₂ = 6.022×10²³ molecules

Therefore,

2.5 mole of CO₂ = 2.5 × 6.022×10²³

2.5 mole of CO₂ = 1.5055×10²⁴ molecules

Thus, 1.5055×10²⁴ molecules are present in 2.5 moles CO₂

7 0
2 years ago
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