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# If the ${{K}_{eq}}$ for a reversible reaction is slightly less than 1, which of the following statements is true?(A)- The formation of products is slightly favored.(B)- The formation of products is greatly favored.(C)- The formation of reactants is slightly favored.(D)- The formation of reactants is greatly favored.(E)- Neither the formation of products or reactants is favored.

Last updated date: 15th Sep 2024
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Hint: The ratio between the product of molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration raised to a power equal to the stoichiometric coefficient in the balanced chemical equation at equilibrium is known as the Equilibrium constant.

Complete Step by step solution:
-The value of reaction quotient at chemical equilibrium, which is a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no more measurable tendency towards further change is given by an equilibrium constant.
-For a given set of reaction conditions, the equilibrium constant is independent of the initial analytical concentrations of the reactants and products in the mixture.
-A knowledge of equilibrium constants is essential for better understanding of many chemical systems, and biochemical processes.
-Let us consider a general reaction:
$aA+bB\rightleftharpoons cC+dD$
This is a reaction in the forward direction. The reaction constant for the forward reaction is given as-
${{k}_{1}}=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$ where $'{{k}_{1}}'$is the equilibrium constant for the forward reaction
The same reaction can be written in a backward direction as-
$cC+dD\rightleftharpoons aA+bB$ where $'{{k}_{2}}'$is the equilibrium constant for backward reaction
The reaction constant or backward reaction is given as-
${{k}_{2}}=\dfrac{{{[A]}^{a}}{{[B]}^{b}}}{{{[C]}^{c}}{{[D]}^{d}}}$
Then, the product of the forward and backward reaction is always 1, that is
${{k}_{1}}\times {{k}_{2}}=1$ ; and ${{k}_{2}}=\dfrac{1}{{{k}_{1}}};{{k}_{1}}=\dfrac{1}{{{k}_{2}}}$
-Therefore, we can say that the value of the equilibrium constant of a reaction is always less than 1.
-Since, ${{K}_{eq}}=\dfrac{{{K}_{forward}}}{{{K}_{backard}}}$ , therefore, ${{K}_{eq}}<1\Rightarrow {{K}_{forward}}<{{K}_{backward}}$
where ${{K}_{forward}}$and ${{K}_{backward}}$are the velocity constant of forward reaction and backward reaction respectively and are independent of the initial concentrations.
-It is now clear that if the rate of the forward reaction is less than the rate of backward reaction, then fewer reactants will react to form products and the equation will shift to the left.

So, the correct answer is option C.

Note: While calculating the value of the equilibrium constant, we need to remember the following points-
(i) Equilibrium constant for a specific reaction at a specific temperature may change if the temperature of the reaction changes.
(ii) Pure solids and pure solvents or pure liquids are not considered in the equilibrium expression.
(iii) The concentrations are usually expressed in molarity, hence the equilibrium constant has unit as $\text{mol }{{\text{L}}^{-1}}$. However, sometimes the equilibrium constant is written without units.
(iv) The reaction must be balanced properly with the coefficients written in the lowest possible integer values to get the correct value for the equilibrium constant.
(v) If the reactants or products are gases, we can write the equilibrium constant in terms of the partial pressure of the gasses and is referred to as ${{K}_{P}}$ to distinguish it from the equilibrium constant.