Question

If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as the reaction proceeds to equilibrium?

Answer 1

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

`COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)`

`K_c=4.74* 10^4`

Concentration at equilibrium:

`[COBr_2]=1.58* 10^(-6)M`

`[Co]=2.78* 10^(-3)M`

`[Br_2]=2.51* 10^(-5)M`

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

`COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)`

The expression for reaction quotient will be :

`Q=([CO][Br_2])/([COBr_2])`

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

`Q=((2.78* 10^(-3))* (2.51* 10^(-5)))/((1.58* 10^(-6)))=4.42* 10^(-2)`

The given equilibrium constant value is,
`K_c=4.74* 10^4`

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When
`Q>K_c` that means product > reactant. So, the reaction is reactant favored.

When
`Q<K_c` that means reactant > product. So, the reaction is product favored.

When
`Q=K_c` that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the
`Q<K_c` that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.