The equilibrium constant K c for the reaction 2NO( g )+O2( g )2NO2( g ) is 6.9×1

Question

The equilibrium constant Kc for the reaction 2NO(g)+O2(g)2NO2(g)
is 6.9×105 at 500 K. A 5.0 L reaction vessel at 500 K was filled with 0.060 mol of NO, 1.0 mol of O2, and 0.80 mol of NO2.

Part A

Is the reaction mixture at equilibrium?

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PART B

In which direction does the net reaction proceed?

In which direction does the net reaction proceed?

Part C

What is the direction of the net reaction if the initial amounts are 5.0×103
mol of NO, 0.20 mol of O2, and 4.0 mol of NO2?

The equilibrium constant Kc for the reaction 2NO(g)+O2(g)2NO2(g)
is 6.9×105 at 500 K. A 5.0 L reaction vessel at 500 K was filled with 0.060 mol of NO, 1.0 mol of O2, and 0.80 mol of NO2.

Part A

Is the reaction mixture at equilibrium?

Is the reaction mixture at equilibrium? This mixture is at equilibrium. This mixture is not at equilibrium.

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PART B

In which direction does the net reaction proceed?

In which direction does the net reaction proceed?

The reaction will proceed to the right to reach equilibrium. The reaction will proceed to the left to reach equilibrium.

Part C

What is the direction of the net reaction if the initial amounts are 5.0×103
mol of NO, 0.20 mol of O2, and 4.0 mol of NO2?

What is the direction of the net reaction if the initial amounts are  
of , 0.20  of , and 4.0  of ? The reaction will proceed to the right to reach equilibrium. The reaction will proceed to the left to reach equilibrium.

Explanation / Answer

[NO] = 0.06

[O2] = 1

[NO2] = 0.80

Since

K = [NO2]^2 / [NO]2 [O2]

Q, the actual coefficient is given as

q =  [NO2]^2 / [NO]2 [O2] = (0.8^2)/(0.06^2)*1) = 177.77

Since Q <<< K, then it favorus reactants; and the mixture is NOT in equilibrium

B)

it goes backward, reverse direction

C)

IF

q =  [NO2]^2 / [NO]2 [O2] = (4^2)/(0.2^2)/(5*10^-3) = 80000

Q = 8*10^4

Q is still less than K, this is in the left/backward reaction; to reach equilibrium

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