Question

Given,${A_{2(g)}} \to 2A(g)$ and for this reaction on increasing $T$ value of ${K_{eq}}$ increases then for this reaction is
A.$\Delta H =$ positive; $\Delta S =$ positive
B.$\Delta H =$ negative; $\Delta S =$ negative
C.$\Delta H =$ positive; $\Delta S =$ negative
D.$\Delta H =$ negative; $\Delta S =$ positive

Answer 1

The relation between enthalpy, entropy and Gibbs free energy is given by the equation $\Delta G = \Delta H - T\Delta S$ . The value of ${K_{eq}}$ can be determined with the help of Gibbs free energy. By knowing the magnitudes of temperature and Gibbs free energy, the nature of enthalpy and entropy can be determined.

Complete step by step answer:
Entropy is a measure of disorder or randomness of the system. It is denoted by $S$ . The absolute value of entropy cannot be calculated. So we calculate the change in entropy occurring during the change of state of system. It is given by the formula-
$\Delta S = \frac{{{q_{rev}}}}{T}$
Where ${q_{rev}}$ is the heat exchanged by the system at isothermal reversible conditions, $T$ is the temperature .
Heat content of a system at a constant pressure is called enthalpy. It is denoted as $\Delta H$ . it is the heat exchange occurring in a system. It is also a state function like entropy. It depends on the initial and final state of the system.
Gibbs free energy is denoted by $\Delta G$. For a reaction taking place at constant temperature and pressure, $\Delta G$ represents that portion of total energy change that is available to do useful work. $\Delta G$ is a state function. It depends on the initial and final state of the system.
The relation between Gibbs free energy $\Delta G$ and equilibrium constant ${K_{eq}}$ of a reaction is given as-
$\Delta G = - 2.303RT\log {K_{eq}}$
Where $R,T$ are the universal gas constant and temperature respectively.
The relation between Gibbs free energy, entropy and enthalpy is given by Gibb’s-Helmholtz equation-
$\Delta G = \Delta H - T\Delta S$
Any process that can occur naturally on Its own is called a spontaneous process. For example, flow of water down a hill, flow of heat from hot body to cold body.
Now each thermodynamic quantity plays an important role in the spontaneity of the process. For example, for a process or reaction if $\Delta G$ is negative then the reaction or process is spontaneous. With the help of Gibb’s-Helmholtz equation and the values of the terms involved in it, we can predict whether the reaction will be spontaneous or nonspontaneous.

This data can be tabulated as follows-
Signs of $\Delta G$ , $\Delta H$ , $\Delta S$ and prediction of spontaneity:

$\Delta H$	$\Delta S$	$\Delta G$	Remarks
$- ve$	$+ ve$	$- ve$	Spontaneous at all temperatures
$- ve$	$- ve$	$- ve$ (at low temperature)$+ ve$ (at high temperature)	SpontaneousNon-spontaneous
$+ ve$	$+ ve$	$+ ve$ (at low temperature)$- ve$ (at high temperature)	Non-spontaneousSpontaneous
$+ ve$	$- ve$	$+ ve$	Non-spontaneous at all temperatures

Now, in the given problem for the reaction, as temperature increases ${K_{eq}}$ increases.
Now we have given the equation, $\Delta G = - 2.303RT\log {K_{eq}}$
From this equation we can see that as ${K_{eq}}$ will increase positively, $\Delta G$ will become more and more negative (it will decrease).
So according to the given condition temperature is high and that temperature the value of $\Delta G$ is negative. So from the table we can see that in such conditions, $\Delta H$ will be positive and $\Delta S$ will also be positive.
In the given reaction,
${A_{2(g)}} \to 2A(g)$
As we can see the number of atoms in a product is increasing, it means that disorder is increasing in the system as the atoms will get dispersed more. So entropy is increasing and $\Delta S$ will be positive.
So the correct option is A.

Note:
$H,G,S$ are all state functions.
If $\Delta G$ is zero, then the system is in equilibrium. The process does not occur.
Any spontaneous process will have a value of $\Delta G$ negative.