
Which reaction is reversible?
(A) $ Cu + \,\,ZnS{O_4}\,\, \to \,\,CuS{O_4}\,\, + \,\,Zn $
(B) $ CuO + \,\,{H_2}S{O_4}\,\, \to \,\,CuS{O_4}\,\, + \,\,{H_2}O $
(C) $ CuO + \,\,{H_2}\,\, \to \,\,Cu\,\, + \,\,{H_2}O $
(D) $ CuS{O_4} \cdot 5{H_2}O\,\, \to \,\,CuS{O_4}\,\, + \,\,5{H_2}O $
Answer
543.9k+ views
Hint: To solve this question, we must first understand the whole concept about reversible reaction. Then we need to assess the main points of the concept which will help us to find which of the given reactions is reversible and then only we can conclude the correct answer.
Complete step by step solution:
Before we move forward with the solution of this given question, let us first understand some basic concepts about Reversible Reaction:
Reversible Reaction: is defined as a chemical reaction where the reactants and the products react together to give the reactants back. In simple words, we can say that it is a reaction involving the simultaneous conversion of reactants to products and vice versa.
The Characteristic of such type of reaction is that the reactants and products are never fully exhausted. Meaning, they are each continuously reacting and being produced. A reversible reaction is indicated or expressed as follows:
$ A + \,\,B\,\, \rightleftarrows \,\,C\,\, + \,\,D $
A and B can react to form C and D and in the reverse reaction, C and D can react to form A and B. The double arrow is the indication that the chemical reaction is reversible.
So after going through all the points stated above we can easily conclude that the only reaction which can be called as Reversible from the given options is:
$ CuS{O_4} \cdot 5{H_2}O\,\, \to \,\,CuS{O_4}\,\, + \,\,5{H_2}O $
In a reversible reaction, reacting molecules in a closed system collide with each other and use the energy to break chemical bonds and form new products. Activation energy is present in the system for the same process to occur with the products.
Therefore the reactants and the products will be in constant equilibrium i.e. there will be continuous interconversion between reactants and product.
So, clearly we can conclude that the correct answer is Option D.
Note:
Reversible reactions may not occur at the same rate in both directions. Nonetheless, an equilibrium condition is achieved. This also depends on the initial concentrations of the reactants and products and the equilibrium constant, K.
Complete step by step solution:
Before we move forward with the solution of this given question, let us first understand some basic concepts about Reversible Reaction:
Reversible Reaction: is defined as a chemical reaction where the reactants and the products react together to give the reactants back. In simple words, we can say that it is a reaction involving the simultaneous conversion of reactants to products and vice versa.
The Characteristic of such type of reaction is that the reactants and products are never fully exhausted. Meaning, they are each continuously reacting and being produced. A reversible reaction is indicated or expressed as follows:
$ A + \,\,B\,\, \rightleftarrows \,\,C\,\, + \,\,D $
A and B can react to form C and D and in the reverse reaction, C and D can react to form A and B. The double arrow is the indication that the chemical reaction is reversible.
So after going through all the points stated above we can easily conclude that the only reaction which can be called as Reversible from the given options is:
$ CuS{O_4} \cdot 5{H_2}O\,\, \to \,\,CuS{O_4}\,\, + \,\,5{H_2}O $
In a reversible reaction, reacting molecules in a closed system collide with each other and use the energy to break chemical bonds and form new products. Activation energy is present in the system for the same process to occur with the products.
Therefore the reactants and the products will be in constant equilibrium i.e. there will be continuous interconversion between reactants and product.
So, clearly we can conclude that the correct answer is Option D.
Note:
Reversible reactions may not occur at the same rate in both directions. Nonetheless, an equilibrium condition is achieved. This also depends on the initial concentrations of the reactants and products and the equilibrium constant, K.
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