Be sure to answer all parts. as an epa scientist studying catalytic converters and urban smog, you want to find kc for the following reaction: 2 no2(g) ⇌ n2(g) + 2 o2(g) kc = ? use the following data to find the unknown kc: 1/2 n2(g) + 1/2 o2(g) ⇌ no(g) kc = 4.8 × 10 −10 2 no2(g) ⇌ 2no(g) + o2(g) kc = 1.1 × 10−5 kc = × 10

The equilibrium constant Kc, for the reaction 2 NO₂(g) ⇌ N2(g) + 2O₂(g) is 4.77 × 10¹³

From the question,

We are to determine the equilibrium constant, Kc for the reaction

2 NO₂(g) ⇌ N2(g) + 2O₂(g)

The equilibrium constant, Kc for this reaction is given by

From the given data,

We have that

¹/₂ N₂(g) + ¹/₂ O₂(g) ⇌ NO(g)            Kc₁ = 4.8 × 10⁻¹⁰

The equilibrium constant, Kc₁, for this reaction is given by

    --------------- (X)

Also,

2NO₂(g) ⇌ 2NO(g) + O₂(g)              Kc₂ = 1.1 × 10⁻⁵

The equilibrium constant, Kc₂, for this reaction is given by

      -------------- (Y)

Now,

From (X)

We can write that

Put this into (Y)

We get

Then,

∴

Recall that,

The equilibrium constant Kc, for the reaction 2NO₂(g) ⇌ N2(g) + 2O₂(g) is given by

∴

From the question,

Kc₁ = 4.8 × 10⁻¹⁰

Kc₂ = 1.1 × 10⁻⁵

∴

∴ Kc = 4.77 × 10¹³

Hence, the equilibrium constant Kc, for the reaction 2NO₂(g) ⇌ N2(g) + 2O₂(g) is 4.77 × 10¹³

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Kc = 4.8x10^13

Explanation:

2 NO2(g) ⇌ N2(g) + 2 O2(g)

(1) 1/2 N2(g) + 1/2 O2(g) ⇌ NO(g) Kc = 4.8 × 10 −10

(2) 2 NO2(g) ⇌ 2NO(g) + O2(g) Kc = 1.1 × 10−5

Using the 2 given reactions you need to form the asked in the first place, in order to do that you multiply one of them by 2 and switch reactants with products and then add them together

(1): 1/2 N2(g) + 1/2 O2(g) ⇌ NO(g) / x2

N2(g) + O2(g) ⇌ 2 NO(g) / switch reactants with products (equivalent to multiply by -1)

2 NO(g) ⇌ N2(g) + O2(g)

(2): 2 NO2(g) ⇌ 2NO(g) + O2(g)

Adding them you get:

2 NO(g) + 2 NO2(g) ⇌ 2NO(g) + O2(g) + O2(g)

After eliminating NO in both sides:

2 NO2(g) ⇌ 2NO(g) + O2(g)

Now that the reaction is formed, everything you multiplied to the reactions you have to do the same in the constant but in the exponent and if you add reactions you multiply each other constants, then:

Kc = (4.8 × 10 −10)^(2*(-1)) *  1.1 × 10−5 = 4.8x10^13

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

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