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 1 month ago

Consider the reaction N2O4 (g) ? 2 NO2 (g). At equilibrium, a 2.00-L reaction vessel contains...

Consider the reaction N2O4 (g) ? 2 NO2 (g). At equilibrium, a 2.00-L reaction vessel contains NO2 at a pressure of 0.269 atm and N2O4 at a pressure of 0.500 atm. The reaction vessel is then compressed to 1.00 L. What will be the pressures of NO2 and N2O4 once equilibrium is re-established?

Consider the reaction: 2 NO2 (g) ⇌ N2O4 (g) The ΔGf° for NO2 (g) = 52...

Consider the reaction: 2 NO2 (g) ⇌ N2O4 (g) The ΔGf° for NO2 (g) = 52 kJ/mol and for N2O4 (g) = 98 kJ/mol at 298 K. Calculate the ΔG at the following values. 0.8 M NO2 (g) & 3.0 M N2O4 (g) 4.0 M NO2 (g) & 0.9 M N2O4 (g) 2.2 M NO2 (g) & 2.4 M N2O4 (g) 2.4 M NO2 (g) & 2.2 M N2O4 (g)

The equilibrium constant (KP) is 0.16 at a particular temperature for the reaction: N2O4(g) ⇌ 2NO2(g)...

The equilibrium constant (KP) is 0.16 at a particular temperature for the reaction: N2O4(g) ⇌ 2NO2(g) Given the following sets of initial conditions, what is the net change that must occur for the reaction to reach equilibrium? Does the reaction shift left to reach equilibrium, does the reaction shift right to reach equilibrium or is the reaction at equilibrium at these initial concentrations so no net change will occur? equilibrium, left, right? PNO2 = 0.154 atm, PN2O4 = 0.148 atm...

Consider the reaction: 2 NO2(g) → N2O4(g) Calculate ΔG (in kJ/mol) at 298°K if the equilibrium...

Consider the reaction: 2 NO2(g) → N2O4(g) Calculate ΔG (in kJ/mol) at 298°K if the equilibrium partial pressures of NO2 and N2O4 are 1.337 atm and 0.657 atm, respectively.

Consider the following equilibrium: N2O4(g)?2NO2(g) Thermodynamic data on these gases are given in Appendix C in...

Consider the following equilibrium: N2O4(g)?2NO2(g) Thermodynamic data on these gases are given in Appendix C in the textbook. You may assume that ?H? and ?S? do not vary with temperature. Part A At what temperature will an equilibrium mixture contain equal amounts of the two gases? Express your answer using four significant figures Part B At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much NO2 as N2O4? Express your answer using four significant...

The degree of dissociation, α, for the following reaction: N2O4(g) <--> 2NO2(g) is 0.655 at 298...

The degree of dissociation, α, for the following reaction: N2O4(g) <--> 2NO2(g) is 0.655 at 298 K and 1.00 bar total pressure. Find K.

For the following reaction, Kc = 0.513 at 500 K. N2O4(g)⇌2NO2(g) If a reaction vessel initially...

For the following reaction, Kc = 0.513 at 500 K. N2O4(g)⇌2NO2(g) If a reaction vessel initially contains an N2O4 concentration of 5.50×10−2 M at 500 K, what are the equilibrium concentrations of N2O4 and NO2 at 500 K? [N2O4], [NO2] = ____ M

The following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M....

The following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M. If an additional 0.10 M NO2 is added to the reaction mixture with no change in volume, what will be the new equilibrium concentration of N2O4? N2O4(g) <> 2 NO2(g) Answer = 0.59 M show work

Consider the reaction: NO2 (g) + CO (g) ---> CO2 (g) + NO (g) The equilibrium...

Consider the reaction: NO2 (g) + CO (g) ---> CO2 (g) + NO (g) The equilibrium constant is at 701 K and 895 K are 2.57 and 567 L mol-1 s-1 so, A. Find the reaction order B. Energy of Activation C. Use part b to find the reaction rate constant at 200 C

The value of Kc for the reaction: N2O4(g) ↔ 2NO2 (g) is 0.21 at 373K. If...

The value of Kc for the reaction: N2O4(g) ↔ 2NO2 (g) is 0.21 at 373K. If a reaction vessel at that temperature initially contains 0.030M NO2 and 0.030M N2O4, what are the concentrations of the two gases at equilibrium? First, the reaction quotient (Q) [page 645 of textbook] must be calculated, and then use the I.C.E table [page 648-650 of the textbook] to determine the equilibrium concentrations.