Respuesta :
Answer:
A. negative ΔG and positive ΔS
Explanation:
According the equation of Gibb's free energy -
∆G = ∆H -T∆S
∆G = is the change in gibb's free energy
∆H = is the change in enthalpy
T = temperature
∆S = is the change in entropy .
And , the sign of the ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,
i.e. ,
if
ΔG < 0 , the reaction is Spontaneous
ΔG > 0 , the reaction is non Spontaneous
ΔG = 0 , the reaction is at equilibrium
From , the given information of the question, the reaction is spontaneous , hence , ΔG is negative.
Entropy -
In a system, the randomness is measured by the term entropy .
Randomness basically refers as a form of energy that can not be used for any work.
The change in entropy is given by amount heat per change in temperature.
When solid is converted to gas entropy increases,
As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.
Hence , from the question,
gaseous product is produced , so entropy increases , so , ΔS is positive.
Therefore ,
The correct option is A. negative ΔG and positive ΔS
Answer:
Option A ( negative ΔG' and positive ΔS') is correct.
Explanation:
If a process is spontaneous or not can be determined by the following equation:
process that occurs at constant temperature and pressure, spontaneity can be determined using the change in Gibbs free energy, which is given by:
ΔG = ΔH - TΔS
→ with ΔH = changes in enthalpy
→ with ΔS = changes in entropy
→ with T = The temperature
To be spontaneous ΔG should be negative, this happens when:
When ΔS > 0 and ΔH < 0, the process is always spontaneous as written.
When ΔS > 0 and ΔH > 0, the process will be spontaneous at high temperatures and non-spontaneous at low temperatures.
When ΔS < 0 and ΔH < 0, the process will be spontaneous at low temperatures and non-spontaneous at high temperatures.
The reaction will not be spontaneous:
When ΔS < 0 and ΔH > 0, the process is never spontaneous, but the reverse process is always spontaneous.
To be spontaneous ΔG should be negative so only option A and B are possible.
The entropy is increasing, so ΔS>0, because a gas (CO2) is being produced and the number of molecules is increasing.
Option A ( negative ΔG' and positive ΔS') is correct.
