Question

For a reaction,$P+Q\to R+S$, the value of $\Delta {{H}^{0}}$ is $-30kJmo{{l}^{-1}}$ and $\Delta S$ is $-100J{{K}^{-1}}mo{{l}^{-1}}$. At what temperature will the reaction be at equilibrium?

Answer 1

The answer to this question is based on the Gibbs – Helmholtz equation which is given by $\Delta G=\Delta H-T\Delta S$ and by substituting the values at the equilibrium conditions, you will get the correct answer.

Complete step by step answer:
- In the classes of physical chemistry, we have come across the concepts of thermodynamics which tells us about various parameters associated with the reaction such as enthalpy of a reaction, entropy of a reaction and also the Gibbs free energy of a reaction.
- We shall now calculate the temperature at which the reaction will be in equilibrium.
- The equation which relates the enthalpy that is$\Delta H$and the entropy that is $\Delta S$ along with the temperature is the Gibbs – Helmholtz equation and this equation is given by,
$\Delta G=\Delta H-T\Delta S$
where, $\Delta G$ is the change in the Gibbs free energy
$\Delta H$ is the change in enthalpy of the system
$T$ is the temperature and
$\Delta S$ is the change in the enthalpy of the system.
- Now, according to the given data, we have
$\Delta {{H}^{0}} = -30kJmo{{l}^{-1}}$, $\Delta S = -100J{{K}^{-1}}mo{{l}^{-1}}$
At equilibrium, the Gibbs free energy value will be ,
$\Delta G=0$
Therefore, the above equation reduces to,
$\Delta H-T\Delta S=0$
- Now, $\Delta {{H}^{0}}=-30kJmo{{l}^{-1}}=30000Jmo{{l}^{-1}}$
Substituting the value in the above equation,
$30000 = T\times 100$
$\Rightarrow T = 300K = {{27}^{0}}C$
Therefore, the correct answer is at temperature $300K$ or ${{27}^{0}}C$, the reaction will be at equilibrium.

Note: The Gibbs – Helmholtz equation is a very important factor because it relates the change in Gibbs energy to its temperature dependence and the position of the equilibrium to change in the enthalpy of the reaction.