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Read the passage given below and answer the following questions from (i) to (v).$\frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2(\text{g})=\text{NH}_3(\text{g});\triangle_\text{r}\text{H}^\ominus=-46.1\text{kJ}\text{mol}^{-1}$
$\frac{1}{2}\text{H}_2(\text{g})+\frac{1}{2}\text{Cl}_2(\text{g})=\text{HCl}(\text{g});\triangle_\text{r}\text{H}^\ominus=-92.32\text{kJ}\text{mol}^{-1}$
$\text{H}_2(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{H}_2\text{O}(\text{l});\triangle_\text{r}\text{H}^\ominus=-285.8\text{kJ}\text{mol}^{-1}$
A spontaneous process is an irreversible process and may only be reversed by some external agency. If we examine the phenomenon like flow of Water down hill or fall of a stone on to the Ground, we find that there is a net decrease in Potential energy in the direction of change. By Analogy, we may be tempted to state that a Chemical reaction is spontaneous in a given Direction, because decrease in energy has Taken place, as in the case of exothermic Reactions. For example: The decrease in enthalpy in passing from Reactants to products may be shown for any Exothermic reaction on an enthalpy diagram. Thus, the postulate that driving force for a Chemical reaction may be due to decrease in Energy sounds ‘reasonable’ as the basis of Evidence so far ! Now let us examine the following reactions:$\frac{1}{2}\text{N}_2(\text{g})+\text{O}_2(\text{g})\rightarrow\text{NO}_2(\text{g});\triangle_\text{r}\text{H}^\ominus=+33.2\text{kJ}\text{mol}^{-1}$
$\text{C}(\text{graphite,s})+2\text{s}(\text{l})\rightarrow\text{CS}_2(\text{l});\triangle_\text{r}\text{H}^\ominus=+128.5\text{kJ}\text{mol}^{-1}$
Therefore, it becomes obvious that while decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases . Whenever heat is added to the system, it increases molecular motions causing increased randomness in the system. Thus heat (q) has randomising influence on the system. Temperature is the measure of average chaotic motion of particles in the system. The entropy change is inversely proportional to the temperature. $\triangle\text{S}$ is related with q and T for a reversible reaction as:$\triangle\text{S}=\frac{\text{q}_\text{rev}}{\text{T}}$
The total entropy change $(\triangle\text{S}_\text{total})$ for the system and surrounding of a spontaneous process is given by $\triangle\text{S}_\text{total}=\triangle\text{S}_\text{system}+\triangle\text{S}_\text{surr}.0$ When a system is in equilibrium, the entropy is maximum, and the change in entropy,$\triangle\text{S}=0.$ We can say that entropy for a spontaneous process increases till it reaches maximum and at equilibrium the change in entropy is zero. Since entropy is a state property, we can calculate the change in entropy of a reversible process by$\triangle\text{S}_{\text{sys}}=\frac{\text{q}_\text{rev,rew}}{\text{T}}$
G = H – TS Gibbs function, G is an extensive property and a state function. The change in Gibbs energy for the system, $\triangle\text{G}_{\text{sys}}$can be written as$\therefore\triangle\text{G}_{\text{sys}}=\triangle\text{H}_{\text{sys}}-\text{T}\triangle\text{S}_{\text{sys}}-\text{S}_\text{sys}\triangle\text{T}$
At constant temperature, $\triangle\text{T}=0$$\triangle\text{G}_{\text{sys}}=\triangle\text{H}_{\text{sys}}-\text{T}\triangle\text{S}_{\text{sys}}$
Usually the subscript ‘system’ is dropped and we simply write this equation as$\triangle\text{G}=\triangle\text{H}–\text{T}\triangle\text{S}$
Thus, Gibbs energy change = enthalpy change – temperature × entropy change, and is referred to as the Gibbs equation, one of the most important equations in chemistry. Here, we have considered both terms together for spontaneity: energy (in terms of $\triangle\text{H}$) and entropy ($\triangle\text{s},$ a measure of disorder) as indicated earlier. Dimensionally if we analyse, we find that $\triangle\text{G}$ has units of energy because, both $\triangle\text{H}$ and the $\text{T}\triangle\text{S}$ are energy terms, since $\text{T}\triangle\text{S}=(\text{K}) (\text{J/K}) = \text{J}.$ Now let us consider how $\triangle\text{G}$ is related to reaction spontaneity. We know, $\triangle\text{S}_\text{systeam}+\triangle\text{S}_\text{surrounding}$ If the system is in thermal equilibrium with The surrounding, then the temperature of the Surrounding is same as that of the system. Also, increase in enthalpy of the surrounding Is equal to decrease in the enthalpy of the System. Therefore, entropy change of Surroundings,$\triangle\text{S}_\text{surr}=\frac{\triangle\text{H}_\text{surr}}{\text{T}}-\frac{\triangle\text{H}_\text{sys}}{\text{T}}$
$\triangle\text{S}_\text{total}=\text{S}_\text{sys}+\Big(-\frac{\triangle\text{H}_\text{sys}}{\text{T}}\Big)$
Rearrangine the above equation:$\text{T}\triangle\text{S}_{\text{total}}=\text{T}\triangle\text{S}_{\text{sys}}-\triangle\text{H}_{\text{sys}}$
For spontaneous process,$\triangle\text{S}_\text{total}>0,$ so
$\text{T}\triangle\text{S}_{\text{sys}}-\triangle\text{H}_{\text{sys}}>\text{O}$
$\Rightarrow-(\triangle\text{H}_{\text{sys}}-\text{T}\triangle\text{S}_{\text{sys}})$
Using equation, the above equation can Be written as:$-\triangle\text{G}>\text{O}$
$\triangle\text{G}=\triangle\text{H}-\text{T}\triangle\text{S},0$
$\triangle\text{H}_\text{sys}$
Is the enthalpy change of a reaction, $\text{T}\triangle\text{S}_\text{sys}$ Is the energy which is not available to Do useful work. So $\triangle\text{G}$ is the net energy Available to do useful work and is thus a Measure of the ‘free energy. For this reason, it Is also known as the free energy of the reaction. $\triangle\text{G}$ gives a criteria for spontaneity at Constant pressure and temperature. If $\triangle\text{G}$ is negative (< 0), the process is b) If $\triangle\text{G}$ is positive (> 0), the process is non Entropy and Second Law of Thermodynamics – For an isolated system the change in energy remains constant. Therefore, increase in entropy in such systems is the natural direction of a spontaneous change. This, in fact is the second law of thermodynamics. Like first law of thermodynamics, second law can also be stated in several ways. The second law of thermodynamics explains why spontaneous exothermic reactions are so common. In exothermic reactions heat released by the reaction increases the disorder of the surroundings and overall entropy change is positive which makes the reaction spontaneous. Absolute Entropy and Third Law of Thermodynamics Molecules of a substance may move in a straight line in any direction, they may spin like a top and the bonds in the molecules may stretch and compress. These motions of the molecule are called translational, rotational and vibrational motion respectively. When temperature of the system rises, these motions become more vigorous and entropy increases. On the other hand when temperature is lowered, the entropy decreases. The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero. This is called third law of thermodynamics. This is so because there is perfect order in a crystal at absolute zero. The statement is confined to pure crystalline solids because theoretical arguments and practical evidences have shown that entropy of solutions and super cooled liquids is not zero at 0K. The importance of the third law lies in the fact that it permits the calculation of absolute values of entropy of pure substance from thermal data alone. For a pure substance, this can be done by summing q T rev increments from 0K to 298K. Standard entropies can be used to calculate standard entropy changes by a Hess’s law type of calculation.$\text{A}+\text{B}\rightleftharpoons\text{C}+\text{D};$ is $\triangle_\text{r}\text{G}=0$
A knowledge of the sign and Magnitude of the free energy change of a Chemical reaction allows: Prediction of the spontaneity of the Chemical reaction. Prediction of the useful work that could Be extracted from it. So far we have considered free energy Changes in irreversible reactions. Let us now Examine the free energy changes in reversible Reactions.‘Reversible’ under strict thermodynamic Sense is a special way of carrying out a Process such that system is at all times in Perfect equilibrium with its surroundings. When applied to a chemical reaction, the Term ‘reversible’ indicates that a given Reaction can proceed in either direction Simultaneously, so that a dynamic Equilibrium is set up. This means that the Reactions in both the directions should proceed with a decrease in free energy, which seems impossible. It is possible only if at equilibrium the free energy of the system is minimum. If it is not, the system would spontaneously change to configuration of lower free energy. So, the criterion for equilibrium Gibbs energy for a reaction in which all reactants and products are in standard state, $\triangle_\text{r}\text{G}=0$ is related to the equilibrium constant of the reaction as follows:$0=\triangle_\text{r}\text{G}^{\ominus}+\text{RT}\text{ln}\text{K}$
or $\triangle_\text{r}\text{G}^{\ominus}=-\text{RT}\text{ln}\text{K}$ or $\triangle_\text{r}\text{G}^{\ominus}=-2.303\text{RT}\log\text{K}$ We also know that$\triangle_\text{r}\text{G}^{\ominus}=\triangle_\text{r}\text{H}^{\ominus}-\text{T}\triangle_\text{r}\text{S}^{\ominus}-\text{RT}\text{ln}\text{K}$
For strongly endothermic reactions, the value of $\triangle_\text{r}\text{H}^\phi$ may be large and positive. In such a case, value of K will be much smaller than 1 and the reaction is unlikely to form much product. In case of exothermic reactions, $\triangle_\text{r}\text{H}^\phi$ is large and negative, and $\triangle_\text{r}\text{G}^\phi$ is likely to be large and negative too. In such cases, K will be much larger than 1. We may expect strongly exothermic reactions to have a large K, and hence can go to near completion. $\triangle_\text{r}\text{G}^\phi$ also depends upon $\triangle_\text{r}\text{S}^\phi,$ if the changes in the entropy of reaction is also taken into account, the value of K or extent of chemical reaction will also be affected, depending upon whether $\triangle_\text{r}\text{S}^\phi$ Is positive or Negative, It is possible to obtain an estimate of $\triangle{\text{G}}^0$ From the measurement of $\triangle{\text{H}}^0$ And $\triangle{\text{S}}^0,$ And then calculate K at any temperature For economic yields of the products. If K is measured directly in the Laboratory, value of $\triangle{\text{G}}^0$ At any other Temperature can be calculated.

1. A spontaneous process is an … process.

1. Irreversible
2. Reversible
3. Partially irreversible
4. Partially reversible

2. $\triangle\text{S}_\text{systeam}+\triangle\text{S}_\text{surrounding}$

1. < 0
2. > 0
3. = 0
4. None of above

3. When a system is in equilibrium, the entropy is maximum, and the change in entropy, $\triangle{\text{S}}.....0.$

1. <
2. >
3. =
4. None of above

4. … does not discriminate between reversible and irreversible process:

1. $\triangle\text{H}$
2. $\triangle\text{S}$
3. $\triangle\text{G}$
4. $\triangle\text{U}$

5. $\text{T}\triangle\text{S}=\ ...$

1. Kg
2. J
3. M
4. lit