# For the reaction at $300K$$A\left( g \right) + B\left( g \right) \to C\left( g \right)$$\Delta H = - 3.0kcal$ ; $\Delta S = - 10.0cal/K$.The value of $\Delta G$ isA) $- 600cal$ B) $- 6600cal$C) $- 6000cal$D) None.

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Hint:We know in thermodynamics, the Gibbs free energy is a thermodynamic potential that can be utilized to ascertain the greatest reversible work that might be performed by a thermodynamic framework at a consistent temperature and weight. The Gibbs free energy is given as,
$\Delta G = \Delta H - T\Delta S$
Where $\Delta G$ change in Gibbs free energy.
The change in enthalpy is given as $\Delta H$.
The change in entropy is given as $\Delta S$.

Given,
The temperature T is $300K$
The change in enthalpy $\Delta H = - 3.0kcal$
The change in entropy $\Delta S = - 10.0cal/K$
Using the Gibbs free energy equation,
$\Delta G = \Delta H - T\Delta S$
We know that $\Delta H = \Delta U + RT\Delta n$ substitute this in above equation we get,
$\Delta G = \Delta U + RT\Delta n - T\Delta S$
On substituting the known values we get,
$\Rightarrow \Delta G = - 3000 + \left( { - 2 \times 300} \right) - \left( { - 10 \times 300} \right)$
On simplifying we get,
$\Rightarrow \Delta G = - 600cal$
Hence option A is correct.

Since the progressions of entropy of synthetic reaction are not estimated promptly, in this manner, entropy isn't commonly utilized as a basis. To deter this trouble, we can utilize G. The indication of $\Delta G$ demonstrates the heading of a substance reaction and decides whether a reaction is spontaneous or not.
-If $\Delta G < 0$ then the reaction is spontaneous toward the path composed (i.e., the reaction is exergonic)
-$\Delta G = 0$ then the reaction is at equilibrium and there is no net change either in forward or turn around course.
-If $\Delta G > 0$ then the reaction isn't spontaneous and the cycle continues immediately in the hold course. To drive such a reaction, we need to have a contribution of free energy (i.e., the reaction is endergonic).