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larisa [96]
3 years ago
7

As an EPA scientist studying catalytic converters and urban smog, you want to find Kc for the following reaction: 2NO2(g) ⇋ N2(g

) + 2 O2(g) Kc = ?. Use the following data to find the unknown Kc (1) ½ N2(g) + 1/2 O2(g) ⇋ NO(g) Kc = 4.8x10-10 (2) 2NO2(g) ⇋ 2NO(g) + O2(g) Kc = 1.1x10-5
Chemistry
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:

Kc = 4.774 * 10¹³

Explanation:

the desired reaction is

2 NO₂(g) ⇋ N₂(g) + 2 O₂(g)

Kc =[N₂]*[O₂]² /[NO₂]²

Since

1/2 N₂(g) + 1/2 O₂(g) ⇋ NO(g)

Kc₁= [NO]/(√[N₂]√[O₂]) →  Kc₁²= [NO]²/([N₂][O₂])

and

2 NO₂(g) ⇋ 2 NO(g) + O₂(g)

Kc₂= [NO]²*[O₂]/[NO₂]² →  1/Kc₂= [NO₂]²/([NO]²[O₂])

then

Kc₁²* (1/Kc₂) = [NO]²/([N₂]*[O₂]) *[NO₂]²/([NO]²[O₂])  = [NO₂]²/([N₂]*[O₂]²) = 1/Kc

Kc₁² /Kc₂ = 1/Kc

Kc= Kc₂/Kc₁² =1.1*10⁻⁵/(4.8*10⁻¹⁰)² = 4.774 * 10¹³

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(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

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Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

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Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

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