Question

A reaction occurs spontaneously if,
A. $T\Delta S < \,\Delta H$ and both $\Delta H$ and $\Delta S$ are $( + )ve$
B. $T\Delta S > \,\Delta H$ and both $\Delta H$ and $\Delta S$ are $( + )ve$
C. $T\Delta S = \,\Delta H$ and both $\Delta H$ and $\Delta S$ are $( + )ve$
D. $T\Delta S > \,\Delta H$ and both $\Delta H$ and $\Delta S$ are $( - )ve$

Answer 1

Hint: Any reaction which proceeds freely up to completion and just needs a start is called a spontaneous process. There are many processes which are non-spontaneous; it means they do not proceed like water climbing up in a tank from the ground. There are some conditions we can see by Gibbs free equation.

Complete step-by-step answer:In the process of spontaneous reaction there is a condition where the entropy of the universe increases. Entropy is called randomness, so if in any reaction there is evolution of gases it will be termed as spontaneous process. Suppose in a reaction olefins are formed which are called as alkenes, so during the reaction if ethene, propene etc. evolved the reaction is called as spontaneous because these are gases and increases the entropy of reaction. This is the theoretical basis on which we can say which is a spontaneous process and which is not.
There is a mathematical relation established known as Gibbs Free energy equation by which we can predict the enthalpy and entropy.
$\Delta G = \Delta H - T\Delta S$
According to the equation for any spontaneous reaction the value of $\Delta G$ should be $( - )ve$ so we can see the relation of other two quantities by the option given and predict which option is correct.
So, in the first option we have given that both $\Delta H$ and $\Delta S$ are $( + )ve$ so if we put the sign of these two intensive properties in the Gibbs free energy relation noting that $T\Delta S < \,\Delta H$ so we a overall positive value. It means as $\Delta G$ is positive here it is not a spontaneous reaction.
Next option says that if $T\Delta S > \,\Delta H$ and that both $\Delta H$ and $\Delta S$ are $( + )ve$ in this case we get the negative value of Gibbs free energy. So this is the main condition for any spontaneous reaction. It is our answer.
In the other two options both $\Delta H$ and $\Delta S$ are $( + )ve$ and $T\Delta S = \,\Delta H$ in that case the $\Delta G$ becomes zero so this cannot be defined as a relation for spontaneity. Lastly we have when $T\Delta S > \,\Delta H$ and both $\Delta H$ and $\Delta S$ are $( - )ve$ in this case if $\Delta S$ becomes negative then the overall value of $T\Delta S$ becomes positive. As given that if here also $T\Delta S > \,\Delta H$ on putting all conditions in the formula we get, a positive Gibbs free energy.

Hence option B is correct.

Note:For spontaneous reaction we can approach through theoretical explanation and also through mathematical explanation. Both have certain rules in the theoretical concept it is a need that during the reaction gas must evolve. So in many reactions for driving the reaction in forward direction this is the driving force. While in Gibb’s energy relation we need to end with a negative value of $\Delta G$ .