ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
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Boiling point.
It is the temperature at which the vapor pressure of the liquid equals the vapor pressure of tha atmosphere that surrounds it, which permit to defeat the attraction among the molecules in the liquid phase and pass to the gas phase.
Yes if we use the resource up before it can renew itself then it could be used up and be depleted or near depletion. The resource has to have time to come back and if we use it up to fast then the resource can not take the time it needs to come back.
Looks like exponential decay. Collision theory is consistent with this model because at higher reactant concentrations we expect the reactants to encounter each other more often (collide) and as the concentration decreases we expect fewer and fewer collision events so the rate of reaction becomes less and less.
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