Question

For a reaction,
$P+Q\to R+S.$ The value of $△H^{o}is-30kJmo{l}^{-1}and△Sis-100J{K}^{-1}mo{l}^{-1}.$ At what temperature the reaction will be at equilibrium?

• A. $27{}^{o}C$
• B. $52{}^{o}C$
• C. $30{}^{o}C$
• D. $45{}^{o}C$

Answer 1

The correct option is A $27{}^{o}C$

We know, $△G=△H-T△S,$
where $\Delta G$ = change in Gibb's free energy
$\Delta H$ = change in enthalpy = $-30,000Jmo{l}^{-1}$
$\Delta S$ = change in entropy=$-100J{K}^{-1}mo{l}^{-1}$
At equilibrium, $△G=0$ for a reaction,
So, $△H=T△S$
$-30000=T\times -100$
$T=300Kor27{}^{o}C$
Hence, the reaction will be at equilibrium at $27{}^{o}C$


Theory:
Gibbs Free Energy : A system parameter that predicts the spontaneity of a chemical reaction . It is the free energy at the constant pressure. Gibbs free energy is also a state function. It is an extensive property that depends on the amount of the system.
$\Delta G=G_{final}-G_{initial}$
Absolute value of Gibbs free energy cannot be determined because Gibbs energy can be given as $G=H-TS$. Gibbs free energy is the function of both enthalpy as well as entropy.
Most of the chemical processes occur in closed or open systems, which undergo both enthalpy & entropy changes.
$G=H-TS$
Differentiating the above equation
$dG=dH-TdS-SdT$

At constant temperature $dT=0$ and for considerable change $dG$ will be $\Delta G$
$\Delta G=\Delta H-T\Delta S$

Physical Interpretation of $\Delta G$ : This is also called non PV work. Magnitude of negative value of $\Delta G$ (process is spontaneous) is the maximum amount of non PV (non expansion) work (useful work eg: electrical work) that can be obtained from the system.

Magnitude of positive value of $\Delta G$ (process is non-spontaneous) is the minimum amount of work (at least amount of work) to be done on the system to make the process occur .