Question

A reaction has a value of $\Delta \text{H=-20Kcal}$ at \[\text{200K}\], the reaction is spontaneous, below this temperature, it is not. The values $\text{ }\!\!\Delta\!\!\text{ G}$ and $\text{ }\!\!\Delta\!\!\text{ S}$ at \[\text{400K}\] are, respectively:
(A)\[\text{10Kcal,-0}\text{.1cal }{{\text{K}}^{\text{-1}}}\]
(B) \[\text{-10Kcal,-100cal }{{\text{K}}^{\text{-1}}}\]
(C) \[\text{0Kcal,10}\text{.0cal }{{\text{K}}^{\text{-1}}}\]
(D) \[\text{20Kcal,-100cal }{{\text{K}}^{\text{-1}}}\]

Answer 1

Hint: At a constant temperature T and pressure P the Gibbs free energy G is defined as the difference between the enthalpy H and the internal energy Q or \[\text{T }\!\!\Delta\!\!\text{ S}\]of the system. For spontaneous reactions, Gibbs free energy for a reaction is equal to zero.

Complete step by step solution:
The Gibbs free energy for a reaction at constant pressure P and temperature T is given by the following equation,
$\text{ }\!\!\Delta\!\!\text{ G= }\!\!\Delta\!\!\text{ H-T }\!\!\Delta\!\!\text{ S}$
Where H is enthalpy, S is entropy and T is the temperature in kelvin. It is represented as the G.
​Here is a problem, we are given with
$\Delta \text{H=-20Kcal}$
We know the relation of Gibbs free energy with enthalpy and entropy as:
$\text{ }\!\!\Delta\!\!\text{ G= }\!\!\Delta\!\!\text{ H-T }\!\!\Delta\!\!\text{ S}$ (1)
From the problem, the reaction is not spontaneous below \[\text{200K}\]temperature. The reaction is at the equilibrium ate \[\text{200K}\]
Therefore, $\text{ }\!\!\Delta\!\!\text{ G=0}$
Let’s substitute these values in (1) and rearrange the equation for entropy S.
$\Delta \text{G= }\!\!\Delta\!\!\text{ H-T }\!\!\Delta\!\!\text{ S}$
$\text{0= }\!\!\Delta\!\!\text{ H-T }\!\!\Delta\!\!\text{ S}$
$\text{ }\!\!\Delta\!\!\text{ H=T }\!\!\Delta\!\!\text{ S}$
$\text{-20000=(200)S}$
$\text{S=}\frac{\text{-20000}}{\text{200}}\text{=100cal/K}$
$\text{S=0}\text{.1Kcal/K}$
Let’s find out the Gibbs free energy at $\text{400K}$
\[\begin{align}
  & \text{At, 400K }\!\!~\!\!\text{ } \\
 & \text{ }\!\!\Delta\!\!\text{ G=(-20+400 }\!\!\times\!\!\text{ 0}\text{.1)Kcal} \\
\end{align}\]
\[\Delta G=20Kcal\]
\[\begin{align}
  & \text{So, }\!\!~\!\!\text{ } \\
 & \text{ }\!\!\Delta\!\!\text{ G=20Kcal }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\!\!\Delta\!\!\text{ S=-100cal/K or 0}\text{.1Kcal/K} \\
\end{align}\]

Hence, (D) is the correct option.

Additional information:
-Gibbs free energy G is used to determine the spontaneity of a process. By spontaneity, we mean that the progress of reaction with or without using external energy.
-We will use the signs to figure out whether the reaction is spontaneous in forward or backward direction and at the equilibrium.
1) When $\text{ }\!\!\Delta\!\!\text{ G0}$i.e. When G is less than zero or negative the process is exergonic and proceeds spontaneously in forward direction to form the product.
2) When $\text{ }\!\!\Delta\!\!\text{ G0}$ then the G is greater than zero or positive .This denotes the endergonic or non-spontaneous reaction. The reaction goes in backward direction to form reactants
3) When $\text{ }\!\!\Delta\!\!\text{ G=0}$the reaction is said to be in equilibrium. The concentration of reactants and products are the same.

Note: Pay an extra close attention to the units of the quantities while calculating the $\text{ }\!\!\Delta\!\!\text{ G}$form $\text{ }\!\!\Delta\!\!\text{ H}$and $\Delta S$.The $\text{ }\!\!\Delta\!\!\text{ H}$is usually given as the$\frac{\text{KJ}}{\text{mol}}$for a reaction and $\Delta S$is given the $\frac{\text{J}}{\text{mol K}}$.There is a difference of 1000 times in units.