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Question: For the reaction at \(300K\) \(A\left( g \right) + B\left( g \right) \to C\left( g \right)\) \(\...

For the reaction at 300K300K
A(g)+B(g)→C(g)A\left( g \right) + B\left( g \right) \to C\left( g \right)
ΔH=−3.0kcal\Delta H = - 3.0kcal ; ΔS=−10.0cal/K\Delta S = - 10.0cal/K.The value of ΔG\Delta G is
A) −600cal - 600cal
B) −6600cal - 6600cal
C) −6000cal - 6000cal
D) None.

Explanation

Solution

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,
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
Where ΔG\Delta G change in Gibbs free energy.
The change in enthalpy is given as ΔH\Delta H.
The change in entropy is given as ΔS\Delta S.

Complete step by step answer:
Given,
The temperature T is 300K300K
The change in enthalpy ΔH=−3.0kcal\Delta H = - 3.0kcal
The change in entropy ΔS=−10.0cal/K\Delta S = - 10.0cal/K
Using the Gibbs free energy equation,
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
We know that ΔH=ΔU+RTΔn\Delta H = \Delta U + RT\Delta n substitute this in above equation we get,
ΔG=ΔU+RTΔn−TΔS\Delta G = \Delta U + RT\Delta n - T\Delta S
On substituting the known values we get,
⇒ΔG=−3000+(−2×300)−(−10×300)\Rightarrow \Delta G = - 3000 + \left( { - 2 \times 300} \right) - \left( { - 10 \times 300} \right)
On simplifying we get,
⇒ΔG=−600cal\Rightarrow \Delta G = - 600cal
Hence option A is correct.

Additional information:
We have to remember that the spontaneous - is a reaction that is consider to be normal since it is a reaction that happens without anyone else with no outside activity towards it. Non spontaneous - needs consistent outer energy applied to it all together for the cycle to proceed and once you stop the outside activity the cycle will stop. When comprehending for the condition, in the event that difference in G is negative, at that point it's spontaneous. On the off chance that difference in G on the off chance that positive, at that point it's non spontaneous. The image that is ordinarily utilized for free energy can be all the more appropriately considered as "standard free energy change.
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 ΔG\Delta G demonstrates the heading of a substance reaction and decides whether a reaction is spontaneous or not.

Note:
-If ΔG<0\Delta G < 0 then the reaction is spontaneous toward the path composed (i.e., the reaction is exergonic)
-ΔG=0\Delta G = 0 then the reaction is at equilibrium and there is no net change either in forward or turn around course.
-If ΔG>0\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).