Question

The equilibrium constant (\[{{K}_{c}}\]) for the reaction \[HA+B\rightleftharpoons B{{H}^{+}}+{{A}^{-}}\] is \[100\]. If the rate constant for the forward reaction is\[{{10}^{5}}\], then the rate constant for the backward reaction is_________.
(A) \[{{10}^{7}}\]
(B) \[{{10}^{3}}\]
(C) \[{{10}^{-3}}\]
(D) \[{{10}^{-5}}\]

Answer 1

Hint: The concentrations of the products change with the concentration of the reactants according to a mathematical relationship known as the equilibrium constant (\[{{K}_{eq}}\]).

Complete Step by Step Solution:
When a chemical process reaches equilibrium, the equilibrium constant (often represented by the letter\[K\]) sheds light on the interaction between the reactants and products. For instance, the ratio of the concentration of the products to the concentration of the reactants, each raised to their respective stoichiometric coefficients, can be used to establish the equilibrium constant of concentration (denoted by\[{{K}_{c}}\]) of a chemical reaction at equilibrium. It is significant to remember that there are several kinds of equilibrium constants that establish relationships between the reactants and products of equilibrium reactions in terms of various units.

The ratio of reactant to product amounts that is used to predict chemical behaviour is the definition of the equilibrium constant for a chemical process.

Any chemical change would be considered a forward reaction if the reactants reacted to produce the products on the right side of the arrow. Any forward movement would shift to the right side, where the formation of products or products occurs. The rate of forward reaction is denoted by \[{{r}_{f}}\] and the equilibrium constant for the forward reaction\[{{K}_{f}}\]. A backward reaction is a reversible reaction that proceeds from right to left and produces reactants from products. The rate of backward reaction is denoted by \[{{r}_{b}}\]and the equilibrium constant for the backward reaction is\[{{K}_{b}}\].
At equilibrium, the rate of the forward reaction = the rate of the backward reaction
That is. \[{{r}_{f}}=r{}_{b}\]Or, \[{{k}_{f}}\times \alpha \times {{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}={{k}_{b}}\times \alpha \times {{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}\]

At a particular temperature, the rate constants are constant. The ratio of the rate constant of forward reaction to the rate constant of backward reaction should be a constant and is called an equilibrium constant (\[{{K}_{eq}}\]).
\[{{K}_{eq}}={{K}_{f}}/{{K}_{b}}={{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}/{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}={{K}_{c}}\]
Where \[{{K}_{c}}\]indicates the equilibrium constant measured in moles per litre.
According to our question,
Given that \[{{K}_{c}}\]=\[100\]
\[{{K}_{f}}={{10}^{5}}\]
Formula: \[{{K}_{c}}={{K}_{f}}/{{K}_{b}}\]
\[100={{10}^{5}}/{{K}_{b}}\]
\[{{K}_{b}}={{10}^{5}}/100\]
\[{{K}_{b}}={{10}^{3}}\]
The correct option is B.

Note: For computing the, separate formulas are used for gases and liquids, respectively. When temperature and concentration are perfect, they have no impact on the calculation, but when they are imperfect, they have a significant impact.