Consider the reaction below: a (aq) ↔ b (aq) kc = 2.36 if the reaction is started by placing 0.134 mol of a into 250.0 ml of solution, what will the concentration of a be at equilibrium?

The equilibrium reaction is as follows:

The initial number of moles of a (aq) and volume of solution is given 0.134 mol and 250 mL.

First calculate the molarity of solution which is defined as number of moles of solute in 1 L of solution.

Putting the values,

Thus, initial concentration of a(aq) will be 0.536 M.

At equilibrium, let the change in concentration be x, thus, concentration of a(aq) will be 0.536-x and that of b(aq) will be x.

The expression for for the equilibrium reaction is:

Here [a] and [b] are equilibrium concentration of reactant a and product b respectively.

Putting the values,

On rearranging,

Now, equilibrium concentration of a (aq) will be:

Therefore, equilibrium concentration of a(aq) is 0.16 M

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

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