Question

If a process is both endothermic and spontaneous then
A. $\Delta S > 0$
B. $\Delta S < 0$
C. $\Delta H < 0$
D. $\Delta G > 0$

Answer 1

Hint : Entropy is a measured physical attribute that is most usually linked with a condition of disorder, unpredictability, or uncertainty. The phrase and idea are utilised in a wide range of domains, from classical thermodynamics, where it was originally discovered, through statistical physics' microscopic description of nature, to information theory's principles. It has a wide range of applications in chemistry and physics, biological systems and their relationships to life, cosmology, economics, sociology, weather science, climate change, and information systems, including telecommunications.

Complete Step By Step Answer:
A spontaneous process in thermodynamics is one that happens without any external input to the system. The time-evolution of a system in which it releases free energy and travels to a lower, more thermodynamically stable energy state is a more technical definition (closer to thermodynamic equilibrium). The sign convention for free energy change is the same as for thermodynamic measurements, with a release of free energy from the system resulting in a negative change in the system's free energy and a positive change in the free energy of the surroundings.
Reactions that happen on their own. When reactions result in a drop in enthalpy and a rise in entropy, they are considered beneficial. During a combustion reaction, the system's entropy grows. Combustion processes are spontaneous because of the combination of energy loss and entropy rise. An endothermic reaction is one in which the energy required to break bonds in the reactants is greater than the energy released when new bonds are formed in the products. An endothermic process requires a steady supply of energy, which is commonly in the form of heat. The free energy change is always positive and the reaction is never spontaneous if the reaction is endothermic ( H positive) and the entropy change S is negative (reduced disorder).
$\Delta S > 0$
Hence option a is correct.

Note :
The free energy is calculated differently depending on the nature of the operation. When evaluating processes that occur under constant pressure and temperature settings, for example, the Gibbs free energy change is employed, but the Helmholtz free energy change is utilised when examining processes that occur under constant volume and temperature circumstances. Temperature, pressure, and volume can affect the value and even the sign of both free energy changes.