Question

In: Chemistry

Consider the following reaction. 2NO2(g)⇌N2O4(g) When the system is at equilibrium, it contains NO2 at a...

Consider the following reaction. 2NO2(g)⇌N2O4(g) When the system is at equilibrium, it contains NO2 at a pressure of 0.722 atm, and N2O4 at a pressure of 0.0521 atm. The volume of the container is then reduced to half its original volume. What is the pressure of each gas after equilibrium is reestablished?

PNO2= ?? atm

PN2O4= ?? atm

Expert Solution

P₁V₁ = P₂V₂ or P₁/P₂ = V₂/V₁ when the volume is reduced to half i.e V₂ = 0.5 V₁ , P₁ / P₂ = 0.5 = 1/2

P₂ = 2 P₁

So when the volume is reduced to half, the pressure will be doubled and it is assumed under isothermal condition.

The equation is

2 NO₂ (g) N₂O₄ (g) Total P

equillibrium pressure 0.722 atm 0.0521 atm 0.722 + 0.0521 = 0.7741 atm

Total Pressure = P_NO2 + P_N2O4

When reducing the volume to half

the total pressure is doubled A atm B atm 2 x 0.7741 = 1.5482 atm

A + B = 1.5482 atm

Where A and B are the new equillibrium pressure of NO₂ and N₂O₄ respectively

K_p of the reaction at the given equillibrium pressures = [P_N2O4] / [P_NO2] ²

= 0.0521 / 0.722² atm^-1

= 0.1 atm^-1

K_p under the new equillibrium conditions is given by

K_p = B / A² but K_p = 0.1 atm^-1

therefore 0.1 = B / A² but A + B = 1.5482 atm and hencer B = [1.5482 - A] atm

0.1 = [1.5482 -A]/ A²

01 A² = 1.5482 - A

or 0.1 A² + A - 1.5482 = 0 This is quadratic equation in A can be solved

using the standard formula

x = [ - b (b² - 4 ac) ] / 2a

A =[ -1 ( 1 - 4 (0.1)(-1.5482)] / 0.2

The pressure cannot be negative and hence we take only the positive value of .

A =[ -1 + (1.61928) ] /0.2

= [ -1 + 1.2725] / 0.2

= 0.2725/0.2 = 1.3625 atm therefore B = 1.5482 - 1.3625 = 0.1857 atm

P_NO2 = 1.3625 atm

P_N2O4 = 0.1857 atm

queen_honey_blossom answered 2 hours ago

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