Question

Исходные концентрации А и В в системе 2А+В=2АВ были соответственно равны 0,04 и 0,02 моль/л. К моменту наступления равновесия концентрация А стала равной 0,01 моль/л. Вычислите равновесные концентрации всех остальных реагирующих веществ.

Answer 1

C(A)(Концентрация прореагировавшего вещества A)=0.04-0.01=0.03 Моль/л 
Если судить по уравнению реакции то C(B)(Концентрация прореагировавшего вещества B)= 0.03/2=0.015 моль/л а значит равновесная концентрация вещеcтва B =0.02-0.015=0.005 моль/л
Если продукт в вашей реакции правильно написан, то есть 2AB а не A2B, а второй вариант более вероятен тогда: Если 2AB то Cравн=C(вещества A прореаг)=0.03 Если A2B то Сравн=С(прореагировавшего вещества B)=0.015

Answer 2

Равновесная концентрация B равна нулю.

Answer 3

Calculation of Equilibrium Concentrations

To calculate the equilibrium concentrations of all the reacting substances in the system 2A + B = 2AB, we need to use the given initial concentrations and the information about the change in concentration of A.

The initial concentrations of A and B were given as 0.04 mol/L and 0.02 mol/L, respectively. The concentration of A at equilibrium is given as 0.01 mol/L.

To find the equilibrium concentrations of A, B, and AB, we can use the stoichiometry of the reaction and the concept of equilibrium constant.

The balanced equation for the reaction is: 2A + B = 2AB

Let's assume the equilibrium concentrations of A, B, and AB are [A], [B], and [AB], respectively.

According to the stoichiometry of the reaction, the change in concentration of A is -0.03 mol/L (from 0.04 mol/L to 0.01 mol/L). Since 2 moles of A react to form 2 moles of AB, the change in concentration of AB is +0.03 mol/L.

Using the given information and the stoichiometry of the reaction, we can set up the following equations:

[A] = 0.01 mol/L [B] = 0.02 mol/L [AB] = 0.03 mol/L

Therefore, the equilibrium concentrations of A, B, and AB are 0.01 mol/L, 0.02 mol/L, and 0.03 mol/L, respectively.

Equilibrium Concentrations: - [A] = 0.01 mol/L - [B] = 0.02 mol/L - [AB] = 0.03 mol/L

Please note that the given information does not provide enough details to calculate the equilibrium constant for the reaction.