Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are...

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Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called NOx) that are of interest to atmospheric

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Answer 1

Answer #1

Solution :-

Balanced reaction equation

2NO2(g) ---- > N2O4(g)

Initial pressure of NO2 = 1.6 atm

Equilibrium pressure of NO2 = 0.61 atm

Lets set up the ICE table

2NO2(g) ---- > N2O4(g)

1.6 atm 0

-2x +x

1.6-2x=0.61 x

Lets calculate the value of x

1.6atm - 2x=0.61 atm

(1.6 atm – 0.61 atm)/2=x

0.495 atm = x

Therefore the equilibrium pressure of the N2O4 is 0.495 atm

Now lets calculate the equilibrium constant

Kp=[N2O4]/[NO2]^2

=[0.495]/[0.61]^2

= 1.33

Therefore the equilibrium constant Kp is 1.33

Now lets calculate the pressure of N2O4 after second equilibrium reached when 0.53 atm of NO2 is added to the container

Therefore pressure of the NO2 gas is (0.61 atm+0.53 atm=1.14 atm)

Lets set up ICE table

2NO2(g) ---- > N2O4(g)

1.14 atm 0.495 atm

-2x +x

1.14-2x 0.495 + x

Now lets calculate the value of x using the Kp equation

Kp=[N2O4]/[NO2]^2

1.33 =[0.495+x]/[1.14-2x]^2

1.33=0.495+x /(4x^2-4.56x+1.2996)

Multiply both sides by (4x^2-4.56x+1.2996)

1.33*(4x^2-4.56x+1.2996)= 0.495+x /(4x^2-4.56x+1.2996)*(4x^2-4.56x+1.2996)

5.32x^2-6.0648x+1.728468=x+0.495

Subtract x+0.495 from both sides

5.32x^2-6.0648x+1.728468 – (x+0.495)=x+0.495 - x+0.495

5.32x^2 - 7.0648x + 1.233468=0

Using the quadratic equation we can get the value of x as follows

X=0.207 atm

Now lets find the equilibrium pressure of the N2O4

P[N2O4]=0.495+x

=0.495 atm + 0.207 atm

=0.70 atm

Therefore the pressure of the dinitrogen tetraoxide after second equilibrium is 0.70 atm.

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