Statement-1 : The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.Statement-2 : The value of entropy change +S of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of +Go becomes more on negative side.a)Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1.b)Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1.c)Statement- 1 is True, Statement- 2 is False.d)Statement- 1 is False, Statement- 2 is True.Correct answer is option 'A'. Can you explain this answer?

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Statement-1 : The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.
Statement-2 : The value of entropy change +S of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of +Go becomes more on negative side.

a)
Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1.
b)
Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1.
c)
Statement- 1 is True, Statement- 2 is False.
d)
Statement- 1 is False, Statement- 2 is True.

Correct answer is option 'A'. Can you explain this answer?

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Statement 1: The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.

Statement 2: The value of entropy change S of the reduction process is more on the negative side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of Go becomes more on the negative side.

Explanation:

Statement 1: The reduction of a metal oxide involves the removal of oxygen from the metal oxide to produce the metal. This process requires energy in the form of heat. The reduction is easier if the metal formed is in a liquid state at the temperature of reduction because in the liquid state, the metal particles have greater mobility and can come into contact with the metal oxide more easily, enhancing the reaction rate. Therefore, Statement 1 is true.

Statement 2: The entropy change (ΔS) is a measure of the disorder or randomness of a system. In a reduction process, the entropy change is related to the physical states of the reactants and products. When the metal formed is in a liquid state and the metal oxide being reduced is in a solid state, the entropy change is more on the negative side. This is because the conversion of a solid into a liquid state is accompanied by an increase in entropy. On the other hand, the conversion of a solid metal oxide into a liquid metal and gaseous oxygen is accompanied by a decrease in entropy. As a result, the overall entropy change for the reduction process is negative.

The change in Gibbs free energy (ΔG) is related to the entropy change (ΔS) and the temperature (T) through the equation ΔG = ΔH - TΔS, where ΔH is the enthalpy change. When the value of ΔS is more on the negative side, the value of ΔG becomes more on the negative side. A more negative value of ΔG indicates a more favorable and spontaneous reaction. Therefore, Statement 2 is true.

Explanation of the relationship between Statement 1 and Statement 2: Statement 2 provides a correct explanation for Statement 1. The increased mobility of metal particles in the liquid state facilitates the reduction process, leading to a more negative entropy change and a more negative value of ΔG. Therefore, the reduction of a metal oxide is easier when the metal formed is in a liquid state, and the explanation for this lies in the more negative entropy change associated with this state. Hence, option A is the correct answer.

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Statement-1 : The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.Statement-2 : The value of entropy change +S of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of +Go becomes more on negative side.a)Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1.b)Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1.c)Statement- 1 is True, Statement- 2 is False.d)Statement- 1 is False, Statement- 2 is True.Correct answer is option 'A'. Can you explain this answer?

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Statement-1 : The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.Statement-2 : The value of entropy change +S of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of +Go becomes more on negative side.a)Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1.b)Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1.c)Statement- 1 is True, Statement- 2 is False.d)Statement- 1 is False, Statement- 2 is True.Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Statement-1 : The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.Statement-2 : The value of entropy change +S of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of +Go becomes more on negative side.a)Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1.b)Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1.c)Statement- 1 is True, Statement- 2 is False.d)Statement- 1 is False, Statement- 2 is True.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Statement-1 : The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.Statement-2 : The value of entropy change +S of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of +Go becomes more on negative side.a)Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1.b)Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1.c)Statement- 1 is True, Statement- 2 is False.d)Statement- 1 is False, Statement- 2 is True.Correct answer is option 'A'. Can you explain this answer?.

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