Question

Select correct statement:
A) In the decomposition of an oxide, into oxygen and gaseous metal, entropy increases.
B) Decomposition of an oxide is an endothermic change.
C) To make $\Delta {{\text{G}}^ \circ }$ negative, temperature should be high enough so that ${\text{T}}\Delta {{\text{S}}^ \circ } > \Delta {{\text{H}}^ \circ }$.
D) All of the above are correct statements.

Answer 1

Hint: To solve this verify each statement carefully. The measure of randomness or disordered distribution is known as entropy. The randomness is always higher in a gaseous state. More the number of gaseous molecules higher is the entropy.

Complete step-by-step answer:
A) When an oxide of any metal is heated, the oxide decomposes to form pure metal and oxygen gas. The reaction is as follows:
${\text{MO}}\left( {\text{s}} \right) \to {\text{M}}\left( {\text{s}} \right) + {{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
We know that entropy is the measure of randomness. The entropy of solids is very low because in solids the atoms or ions or molecules have an ordered arrangement. The entropy of gases is higher because the constituents of gases are not in ordered arrangement.
In the reaction, a solid reactant is converted to two gaseous products. Thus, the entropy increases as the reaction proceeds.
Thus, the statement ‘in the decomposition of an oxide, into oxygen and gaseous metal, entropy increases’ is correct.
B) When an oxide of any metal is heated, the oxide decomposes to form pure metal and oxygen gas. The reaction is as follows:
${\text{MO}}\left( {\text{s}} \right) \to {\text{M}}\left( {\text{s}} \right) + {{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
In the reaction, we can see that the bond between the metal and oxygen breaks. A large amount of energy is needed to break this bond. This energy is absorbed from the surrounding.
We know that the reactions in which energy is absorbed are known as endothermic reactions. Thus, decomposition of an oxide is an endothermic reaction.
Thus, the statement ‘decomposition of an oxide is an endothermic change’ is correct.
C) We know the expression that gives the relation between free energy, entropy and enthalpy is as follows:
$\Delta {{\text{G}}^ \circ } = \Delta {{\text{H}}^ \circ } - {\text{T}}\Delta {{\text{S}}^ \circ }$
Where, $\Delta {{\text{G}}^ \circ }$ is the change in Gibb’s free energy,
 $\Delta {{\text{H}}^ \circ }$ is the change in enthalpy,
 $T$ is the temperature,
$\Delta {{\text{S}}^ \circ }$ is the change in entropy.
When temperature is high ${\text{T}}\Delta {{\text{S}}^ \circ } > \Delta {{\text{H}}^ \circ }$. And thus, $\Delta {{\text{G}}^ \circ }$ becomes negative.
Thus, the statement ‘to make $\Delta {{\text{G}}^ \circ }$ negative, temperature should be high enough so that ${\text{T}}\Delta {{\text{S}}^ \circ } > \Delta {{\text{H}}^ \circ }$’ is correct.

Thus, the correct option is (D) all of the above are correct statements.

Note:For an endothermic process, the value of $\Delta H$ is positive. In an endothermic process, heat is absorbed and the system feels cold. A negative value of $\Delta H$ indicates that the reaction is an exothermic process. In an exothermic process, heat is released or given out