Calculate The Equilibrium Concentrations Of N2O4and NO2 At ~ BestNews

Thursday, April 22, 2021

Calculate The Equilibrium Concentrations Of N2O4and NO2 At

April 22, 2021 / by NewsGeek / with No comments /

Dinitrogentetraoxide partially decomposes according to the following equilibrium: N2O4 (g) ⇌ 2NO2 (g) A 1.00 L flask is charged with 0.0400 mol of N2O4 at 373 K. An equilibrium is established after some time and only 0.0055 mol of N2O4 remains. What is KC for this reaction? Select one: a. Kc = 0.866 b. Kc = 0.216 c. Kc = 12.5 d.[2ΔS f (NO2 (g))] - [1ΔS f (N2O4 (g))] [2(239.95)] - [1(304.18)] = 175.72 J/K 175.72 J/K (increase in entropy)Answer to A chemist measures the energy change ΔH during the following reaction: 2NO2(g)→N2O4(g) ΔH=−55.3kJ Use the informa...The reaction N2O4 (g) ↔ 2NO2 (g) is an equilibrium reaction at some temperature with an equilibrium constant K = 4.00 x 10−7. If the intial condition is pure N2O4 (g) at a concentration of 1.20...The reaction N2O4 (g) ↔ 2NO2 (g) is an equilibrium reaction at some temperature with an equilibrium constant K = 4.00 x 10−7. If the intial condition is pure N2O4 (g) at a concentration of 1.73...

N2O4 (g) → 2 NO2 (g) - Stoichiometry - Enthalpy • Entropy

2NO2(g) ⇌ N2O4(g) Keq= [N2O4(g)] / [NO2(g)]^2. substituting the given data we get . 1.15 = [N2O4(g)] / ( 0.50)^2 [N2O4(g)] = 1.15 * ( 0.50)^2 = 0.2875 moles / literAt a particular temperature, Kp = 0.370 for the reaction N2O4(g) 2NO2(g) A flask containing only N2O4(g) at an initial pressure of 3.20 atm is allowed to reach equilibrium. Calculate the total pressure in this flask at equilibrium. I need some serious help..I'v tried everything with this problem...this is what i tried before and it doesnt work. Kp for the reaction can be written as Kp= ((pNO22 NO 2 (g) <=> N 2 O 4 (g) The standard enthalpy (delta H� = -57.2 kJ) and the entropy (delta S� = -175.83 kJ) of reaction can be calculated from the follow standard-state enthalpies of formation and standard-state entropies.Answer to: Consider the following reversible reaction: N2O4(g) arrow 2NO2(g) a. If we add in more NO2, we will force the reaction further to the....

$N2O4 (g) → 2 NO2 (g) - Stoichiometry - Enthalpy • Entropy$

Solved: A Chemist Measures The Energy Change ΔH During The

Click here👆to get an answer to your question ️ For the reaction 2NO2(g) N2O4(g), Kp/Kc is equal to :Solution for At a particular temperature, Kp = 0.310 for the reaction N2O4(g) 2NO2(g) A flask containing only N2O4(g) at an initial pressure of 3.70 atm…A 1.00 mol sample of N2O4 (g) is placed in a 10.0 L vessel and allowed to reach equilibrium accourding to the equation N2O4(g) 2NO2(g) K = .0004 Calculate the equilibrium concentrations of N2O4(g) and NO2(g) Chemistry. The standard molar Gibbs free energy of formation of NO2 (g) at 298 K is 51.30 kJ · mol −1 and that of N2O4 (g) is 97.82 kJN2O4(g) <--> 2 NO2(g) ∆H = 58.0 kJ A computer animation representing what occurs at the particulate level was available but it is missing. The computer animation needs to be re-built. A YouTube video of this demonstration is availableCalorimetric studies show that the reaction is exothermic. 2NO2(g) N2O4(g) + 14.1 kcal. Based on this information, which one--if any--of the following additional changes would increase the molar concentration at equilibrium of N2O4(g)? decrease the pressure increase the temperature decrease the concentration of NO2(g) stir the reaction mixture

IMO this can be a bit garbled (I'm now not blaming you for that!).

I feel is is how it is meant to be solved.

2NO2(g) → N2O4(g)

Kp = P(merchandise)/P(reactants) = (P(N2O4))/((P(NO2))^2)

Kp = (1.64 atm)/((0.38 atm)^2) = 11.36 atm^-1

Next we want to convert Kp to Kc:

Kp = Kc(RT)^Δngas

Kc = Kp/((RT)^Δngas)

Δngas = Mole of gasoline (product) - Mole of Gas (Reactant) = 1 - 2 = -1

R = 0.08206 L atm Ok^-1 mol^-1

T = 298 Okay

Kc = Kp/((RT)^Δngas) = (11.36 atm^-1)/(((0.08206 L atm K^-1 mol^-1)*(298 Ok))^-1)