Thermodynamics

NEET Chemistry · 93 questions · Page 10 of 10 · Click an option or "Show Solution" to reveal answer

Q91

Which of the following is not correct?

\Delta \mathrm{G}

is positive for a spontaneous reaction

\Delta \mathrm{G}

is positive for a non-spontaneous reaction

\Delta \mathrm{G}

is zero for a reversible reaction

\Delta \mathrm{G}

is negative for a spontaneous reaction

Correct Answer

Option A

Solution

The statement that is not correct among the given options is Option A. Option A states that

\Delta \mathrm{G}

is positive for a spontaneous reaction, which is incorrect.

The criterion for spontaneity in a chemical reaction is based on the Gibbs free energy change (

\Delta \mathrm{G}

) for the process.

A spontaneous reaction is one that occurs without needing continuous input of energy from an external source.

The correct relation is that for a spontaneous reaction,

\Delta \mathrm{G}

is negative (

\Delta \mathrm{G} Option B accurately states that

\Delta \mathrm{G}

is positive for a non-spontaneous reaction. A positive value of

\Delta \mathrm{G}

(

\Delta \mathrm{G} > 0

) indicates that the reaction is not spontaneous under the given conditions and requires external energy to proceed. Option C is correct in stating that

\Delta \mathrm{G}

is zero for a reversible reaction at equilibrium. When a reaction is at equilibrium, it has reached a state where the forward and reverse reactions occur at equal rates, and there is no net change in the composition of the system. At this point, the Gibbs free energy is at its minimum for the given conditions, and

\Delta \mathrm{G} = 0

, indicating that the system is in a state of maximum stability and no further net change can occur without the input or removal of energy. Option D correctly states that

\Delta \mathrm{G}$$ is negative for a spontaneous reaction.

As explained previously, a negative Gibbs free energy change signifies that a process or reaction can proceed spontaneously in the direction written.

Q92

The hydration energies of

K^+

and

Cl^-

are

-x

and

-y

kJ/mol respectively. If the lattice energy of KCl is

-z

kJ/mol, then the heat of solution of KCl is :

x + y + z

z - (x + y)

-z - (x + y)

x - y - z

Correct Answer

Option B

Solution

$\mathrm{KCl}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\Delta \mathrm{H} \text{ sol. }} \mathrm{K}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{\text{(aq) }}^{-1}$

\begin{aligned} \Delta \mathrm{H}_{\mathrm{Sol}^{\mathrm{n}} .} & =\mathrm{L} \cdot \mathrm{E} \cdot+(\mathrm{H} \cdot \mathrm{E})_{\mathrm{K}_{(\mathrm{g})}^{+}}+(\mathrm{HE})_{\mathrm{Cl}_{(\mathrm{g})}^{-1}} \\ & =\mathrm{Z}-\mathrm{x}-\mathrm{y} \\ & =\mathrm{z}-(\mathrm{x}+\mathrm{y}) \end{aligned}

Q93

Let us consider a reversible reaction at temperature, T. In this reaction, both

\Delta \mathrm{H}

and

\Delta \mathrm{S}

were observed to have positive values. If the equilibrium temperature is Te , then the reaction becomes spontaneous at:

\mathrm{Te}>\mathrm{T}

\mathrm{T}>\mathrm{Te}

\mathrm{T}=\mathrm{Te}

\mathrm{Te}=5 \mathrm{~T}

Correct Answer

Option B

Solution

For reaction to be spontaneous according to $2^{\text{nd }}$ law:

\begin{aligned} & \Delta \mathrm{G}\left(\frac{\Delta \mathrm{H}}{\Delta \mathrm{~S}}\right)=\mathrm{T}_{\mathrm{e}} \\ & \Rightarrow \mathrm{~T}>\mathrm{T}_{\mathrm{e}} \end{aligned}

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