Question

For the reversible reaction which k values would indicate that there is more B than A at equilibrium?

Answer 1

Concepts and reason

The concept used to solve this problem is based on the equilibrium constant of a chemical reaction.

The value of the reaction quotient when a reaction reaches equilibrium is called equilibrium constant. It is the ratio of the concentrations of products and concentrations of reactants.

Fundamentals

For a chemical reaction,

$a{\rm{A}} + b{\rm{B}} \to c{\rm{C}} + d{\rm{D}}$

The equilibrium constant is written as follow.

$K = \frac{{{{\left[ {\rm{C}} \right]}^c}{{\left[ {\rm{D}} \right]}^d}}}{{{{\left[ {\rm{A}} \right]}^a}{{\left[ {\rm{B}} \right]}^b}}}$

Here $\left[ {\rm{C}} \right]$ and $\left[ {\rm{D}} \right]$ are concentrations of products. The concentrations of reactants are $\left[ {\rm{A}} \right]$ and$\left[ {\rm{B}} \right]$.

For the reaction,

The equilibrium constant is written as follow.

$K = \frac{{\left[ {\rm{B}} \right]}}{{\left[ {\rm{A}} \right]}}$

The values of K, 0.2 and $9 \times {10^{ - 9}}$ are less than one. Thus, at these values the reactant A is more than product B at equilibrium.

The values of K, 9000 and $9 \times {10^9}$ are more than one. Thus, at these values the product B is more than reactant A at equilibrium.

Ans:

The values of K at which B is more than A are 9000 and $9 \times {10^9}$.