Question

At what temperature would a reaction having $\Delta H = 4K{\text{ cal mo}}{{\text{l}}^{ - 1}}$ and $\Delta S = 10{\text{ cal }}{{\text{k}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ be spontaneous
(A) 400 k
(B) Above 400 k
(C) Below 400 k
(D) Uncertain

Answer 1

Temperature is a fundamental quantity and it measures the degree of how hot or cold a substance is. The unit of temperature is kelvin and its symbol is K. We shall substitute the values in the equation given to calculate the temperature.
$\Delta H = T\Delta S$
Where T is the temperature, H is the enthalpy and S is the entropy.

Complete Step by step solution
The minimum kinetic energy needed for colliding molecules to undergo a chemical reaction is known as the activation energy.
For endothermic reactions, the activation energy of a forward reaction is greater than the activation energy of a backward reaction. For exothermic reactions, the activation energy of a forward reaction is greater than the activation energy of a backward reaction.
$\Delta H$ is the energy difference between the activation energies of a forward reaction and a backward reaction. $\Delta H$ is the enthalpy change. So, enthalpy change is the amount of heat evolved or absorbed by a reaction carried out at constant pressure throughout.
$\Delta S$ is the entropy change. It is the measurement of randomness or disorder of a system. It talks about the spontaneity of a reaction.
A spontaneous reaction is a reaction which favours the formation of products in a given set of conditions like temperature and pressure. A non-spontaneous reaction is a reaction which does not favour the formation of products in a given set of conditions.
To calculate the temperature at which a reaction will be spontaneous and having $\Delta H = 4K{\text{ cal mo}}{{\text{l}}^{ - 1}}{\text{ or 4000 cal mo}}{{\text{l}}^{ - 1}}$ and $\Delta S = 10{\text{ cal }}{{\text{k}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ , the following formula will be applied,
$\Delta H = T\Delta S$
where T is the temperature.
Hence $T = \dfrac{{\Delta H}}{{\Delta S}} = \dfrac{{4000}}{{10}} = 400K$
Therefore, the correct answer is option A, that is, 400 K.

Note
$\Delta H$ = activation energy of forward reaction – activation energy of backward reaction.
So, if activation energy of both forward as well as backward reactions is equal then the difference between them will be zero.
Hence $\Delta H = 0$ . For endothermic reactions $\Delta H$ is positive or greater than zero as activation energy of forward reaction is more than that of backward reaction and for exothermic reactions $\Delta H$ is negative or less than zero as activation energy of backward reaction is more than that of forward reaction.