Consider the following reaction:

2NO(g)+O2(g)→2NO2(g)
Estimate ΔG∘ for this reaction at each of the following temperatures and predict whether or not the reaction will be spontaneous. (Assume that ΔH∘ and ΔS∘ do not change too much within the give temperature range.)
A) 298K
B) 727K
C)859K

The answer for the first is -72.5

To estimate ΔG∘ for the reaction at each temperature, we can use the equation:

ΔG∘ = ΔH∘ - TΔS∘

Where:
ΔG∘ is the standard Gibbs free energy change
ΔH∘ is the standard enthalpy change
ΔS∘ is the standard entropy change
T is the temperature (in Kelvin)

Let's calculate ΔG∘ for each temperature given:

A) At 298K:
We need the values of ΔH∘ and ΔS∘ for the reaction. Let's assume the values are -180 kJ/mol and -200 J/mol*K, respectively.

ΔG∘ (298K) = -180 kJ/mol - (298K)(-200 J/mol*K)
ΔG∘ (298K) = -180 kJ/mol + 59800 J/mol

To convert J to kJ, divide by 1000:
ΔG∘ (298K) = -180 kJ/mol + 59.8 kJ/mol
ΔG∘ (298K) = -120.2 kJ/mol

Since ΔG∘ (298K) is negative, the reaction will be spontaneous at this temperature.

B) At 727K:
Using the same ΔH∘ and ΔS∘ values, we have:

ΔG∘ (727K) = -180 kJ/mol - (727K)(-200 J/mol*K)
ΔG∘ (727K) = -180 kJ/mol + 145400 J/mol
ΔG∘ (727K) = -180 kJ/mol + 145.4 kJ/mol
ΔG∘ (727K) = -34.6 kJ/mol

Since ΔG∘ (727K) is negative, the reaction will also be spontaneous at this temperature.

C) At 859K:
Again, using the same ΔH∘ and ΔS∘ values:

ΔG∘ (859K) = -180 kJ/mol - (859K)(-200 J/mol*K)
ΔG∘ (859K) = -180 kJ/mol + 171800 J/mol
ΔG∘ (859K) = -180 kJ/mol + 171.8 kJ/mol
ΔG∘ (859K) = -8.2 kJ/mol

Since ΔG∘ (859K) is negative, the reaction will be spontaneous at this temperature as well.

In summary:
A) At 298K: ΔG∘ = -120.2 kJ/mol, reaction spontaneous
B) At 727K: ΔG∘ = -34.6 kJ/mol, reaction spontaneous
C) At 859K: ΔG∘ = -8.2 kJ/mol, reaction spontaneous

To estimate ΔG∘ for the given reaction at different temperatures and determine whether the reaction will be spontaneous or not, we will use the Gibbs Free Energy equation:

ΔG∘ = ΔH∘ - TΔS∘

where ΔG∘ is the standard Gibbs Free Energy change, ΔH∘ is the standard enthalpy change, ΔS∘ is the standard entropy change, and T is the temperature in Kelvin.

To proceed, we need the values of ΔH∘ and ΔS∘ for the reaction. These values are usually provided in a reference source, such as a chemistry textbook or table. However, since the values are not given in this case, we will assume them as constant within the given temperature range.

Let's go step by step and estimate ΔG∘ for each temperature:

A) 298K:
In this case, we substitute T = 298K into the equation:
ΔG∘ = ΔH∘ - TΔS∘
We calculate ΔG∘ using the given equation and the values of ΔH∘ and ΔS∘.

B) 727K:
Similarly, for this temperature, we substitute T = 727K into the equation:
ΔG∘ = ΔH∘ - TΔS∘
We calculate ΔG∘ using the given equation and the values of ΔH∘ and ΔS∘.

C) 859K:
Again, for this temperature, we substitute T = 859K into the equation:
ΔG∘ = ΔH∘ - TΔS∘
We calculate ΔG∘ using the given equation and the values of ΔH∘ and ΔS∘.

After calculating ΔG∘ for each temperature, we compare the sign of ΔG∘ to predict whether the reaction will be spontaneous or not. If ΔG∘ is negative, the reaction will be spontaneous. If ΔG∘ is positive, the reaction will not be spontaneous.

dGorxn = (n*dGo formation products) = (n*dGo formation reactants)

dGorxn - is spontaneous.
dGorxn + is not spontaneous.
dGorxn = 0 reaction is at equilibrium.