9) (a) Explain giving reason "The net entropy of the universe tends to increase". (b) For the reaction H_(2(g))+1/2O_(2(g))arrow H_(2)O_((l)) the values of enthalpy change and free energy change are -68.32 and -56.69kcal respectively at 25^circ C Calculate the value of free energy change at 30^circ C (c) Write down the applications of Gibb's Helmholtz equation.

(a) The net entropy of the universe tends to increase because it is a fundamental principle of thermodynamics known as the second law of thermodynamics. Entropy is a measure of the disorder or randomness in a system. When a process occurs, the total entropy of the system and its surroundings always increases. This means that the energy available to do useful work decreases, and the system tends to move towards a state of maximum entropy, which is a state of higher disorder and randomness.<br /><br />(b) To calculate the value of free energy change at $30^{\circ }C$, we can use the following equation:<br /><br />$\Delta G = \Delta H - T \Delta S$<br /><br />where $\Delta G$ is the change in Gibbs free energy, $\Delta H$ is the change in enthalpy, $T$ is the temperature in Kelvin, and $\Delta S$ is the change in entropy.<br /><br />Given that the values of enthalpy change and free energy change are $-68.32$ and $-56.69kcal$ respectively at $25^{\circ }C$, we need to find the value of free energy change at $30^{\circ }C$.<br /><br />First, we need to convert the temperature from Celsius to Kelvin:<br /><br />$T = 30^{\circ }C + 273.15 = 303.15K$<br /><br />Next, we need to find the change in entropy, $\Delta S$. We can do this by rearranging the equation for the change in Gibbs free energy:<br /><br />$\Delta S = \frac{\Delta H - \Delta G}{T}$<br /><br />Substituting the given values:<br /><br />$\Delta S = \frac{-68.32 - (-56.69)}{25 + 273.15} = \frac{-11.63}{298.15} = -0.0391 kcal/K$<br /><br />Now, we can calculate the value of free energy change at $30^{\circ }C$:<br /><br />$\Delta G = \Delta H - T \Delta S = -68.32 - 303.15(-0.0391) = -68.32 + 11.89 = -56.43 kcal$<br /><br />Therefore, the value of free energy change at $30^{\circ }C$ is $-56.43 kcal$.<br /><br />(c) Gibb's Helmholtz equation, also known as the Gibbs free energy equation, has several applications. Some of them include:<br /><br />1. Predicting the spontaneity of a reaction: The sign of the change in Gibbs free energy ($\Delta G$) determines whether a reaction is spontaneous or not. If $\Delta G$ is negative, the reaction is spontaneous, and if $\Delta G$ is positive, the reaction is non-spontaneous.<br /><br />2. Determining the equilibrium constant: The equilibrium constant (K) is related to the change in Gibbs free energy through the following equation:<br /><br />$\Delta G = -RT \ln K$<br /><br />where R is the gas constant and ln is the natural logarithm. This equation allows us to calculate the equilibrium constant if we know the change in Gibbs free energy.<br /><br />3. Calculating the maximum reversible work: The change in Gibbs free energy represents the maximum reversible work that can be done by a system at constant temperature and pressure. This is useful in thermodynamic calculations involving reversible processes.<br /><br />4. Designing chemical reactions: Gibb's Helmholtz equation helps in designing chemical reactions by providing information about the energy changes involved. It allows us to predict the feasibility of a reaction and the conditions under which it will occur.<br /><br />These are just a few examples of the applications of Gibb's Helmholtz equation in various fields of chemistry and thermodynamics.