The following reactions have the indicated equilibrium constants at a particular

Question

The following reactions have the indicated equilibrium constants at a particular temperature:

Determine the values of the equilibrium constants for the following equations at the same temperature:

Be sure to answer all parts. Enter your answers in scientific notation. The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) 2NO(g ) 2N0(g) + O2(g) 2NO2(g ) Determine the values of the equilibrium constants for the following equations at the same Kc = 4.3 × 10-25 K-6.4109 temperature: (a) 4NOg ) N2(g ) + 2NO2(g ) × 10 (b) 4NO2(g) 2N2(g) + 402(g) × 10 (c) 2N0(g) +2NO2(g) 302(g)+2N2(g) × 10

Explanation / Answer

a) Reverse of Eq (1), 2NO <------------> N2 + O2 K1 = 1/Kc = 1/ 4.3 x 10-25

   Eq (2) , 2NO +O2 <------> 2NO2 K2 = 6.4 x 109

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Reverse of Eq (1) + Eq (2)   

2NO + 2NO +O2 <------> N2 + 2NO2 + O2

Cancel O2 on both sides,

4NO <------> N2 + 2NO2

K = K1K2 =  [1/ 4.3 x 10-25] x 6.4 x 109 =1.48 x 1034

b) Reverse of Eq (1), 2NO <------------> N2 + O2 K1 =1/Kc = 1/ 4.3 x 10-25

Multiply with 2 , 4NO <---------> 2N2 + 2O2 ---- Eq (3) K2 = K12 = (1/ 4.3 x 10-25)2

Reverse of Eq (2), 2NO2 <------------> 2NO + O2 K3 =1/Kc = 1/ 6.4 x 109

Multiply with 2 ,   4NO2 <------------> 4NO +2 O2 --- Eq (4) K4 = K32 = (1/ 6.4 x 109)2

4NO <---------> 2N2 + 2O2 ---- Eq (3) K2 = (1/ 4.3 x 10-25)2

4NO2 <------------> 4NO +2 O2 --- Eq (4) K4 = (1/ 6.4 x 109)2

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Eq(3) + Eq(4)   4NO2 <------------> 2N2 +4 O2 ( cancelled 4NO on both sides)

K = K2.K4 = (1/ 4.3 x 10-25)2 x (1/ 6.4 x 109)2 = 1.32 x 1029

c)

Reverse of Eq (1), 2NO <------------> N2 + O2 K1 =1/Kc = 1/ 4.3 x 10-25

Multiply with 2 , 4NO <---------> 2N2 + 2O2 ---- Eq (3) K2 = K12 = (1/ 4.3 x 10-25)2

Reverse of Eq (2), 2NO2 <------------> 2NO + O2 ----- Eq(4) K3 =1/Kc = 1/ 6.4 x 109

Eq (3) + Eq (4) , 2NO2 + 4NO  <------------> 2NO + 2N2 + 3O2

Resultant equation is   2NO2 + 2NO  <------------> 2N2 + 3O2

K =K2.K3 = (1/ 4.3 x 10-25)2 x (1/ 6.4 x 109)= 8.45 x 1038

  K = 8.45 x 1038

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