Question

consider the following system at equilibrium where delta H = 16.1 kj, and Kc = 6.50x10^-3, at 298 K. 2NOBr forwsrd reverse arrows 2NO + Br2

if the VOLUME of the equilibrium system is suddenly increased at constant temperature:

The value of Kc:

increases

decreases

remains thesame

the value of Qc:

is greater than Kc

equal to Kc

is less than Kc

the reaction must:

run in the forward direction to restablish equilibrium

run in the reverse

remain the same. its alrrady at equilibrium

the number of moles of Br2 will:

increase

decrease

remains the same

Answer 1

1. The value of Kc remains same as Kc depicts equilibrium constant which is only changed if the temperature is varied. But here the sudden volume is increased at constant temperature and hence no chnge is Kc is observed.

Kc= [NO]2 [Br2]/ [NOBr]2

2. Qc is less than Kc

Le Chatelier’s principle is used to understand the effect of the variable on the equilibrium constant. This principle states that if any chemical reaction at equilibrium is forced to a change in pressure, temperature, volume or concentration then the equilibrium moves in that direction where the effect of the change is minimized.

The given reaction is as,

Since the volume of the equilibrium system is increased suddenly at a constant temperature, and hence the pressure is decreased the reaction moves in that direction where the number of moles of the gaseous compound is more.

From the given reaction it is evident that,

The moles of the reactant is 2.

The moles of the product is 3.

Since the number of moles is more towards the product side, hence, the reaction moves to the right side to re-establish the equilibrium.

Qc: Reaction Quotient tells about the relative amounts of products and reactants present during a reaction at a particular point in time.The Q value can be compared to the Equilibrium Constant, K, to determine the direction of the reaction that is taking place.

Qc= [NO]2 [Br2]/ [NOBr]2

So, if you decrease the pressure by increasing the volume, the pressure of the gas inside the container will decrease. Now the system will re-establish equilibrium by shifting to side of the reaction with the greatest number of moles of gas. Since the reactant has two moles of gas, and the product has 3 moles of gas. Therefore equilibrium will shift towards the right side i.e. product side.

Dividing by a smaller number will make Q larger and you'll find that after decreasing the pressures Q < K.

3. The reaction must run in forward direction to restablish equilibrium because volume of the equilibrium system is increased suddenly at a constant temperature, and hence the pressure is decreased. Thus the reaction moves in that direction where the number of moles of the gaseous compound is more.

4. Number of moles of Br2 will remain same as changing the volume of the container will not alter the number of moles present in the reaction mixture.