Is the following reaction spontaneous at 25*C?
delta s reaction = -217 J/K*mol
delta H rxn = -1202 KJ
2Mg (s) + O2 (g) -> 2MgO (s)
delta suniverse = delta s system - delta H system / T
i got positive (+) 3.81 x 10 ^ 3 J/K
and my notes says it is spontaneous but i feel like it is non spontaneous
because when delta g is negative and delta h is < tdeltas it should be spontaneous & the opposite of that makes it non spon for an endothermic reaction....i hope this makes sense
my question is why is it spontaneous? thank you in advance
I'm having trouble making sense of it. If DS for the reaction = -217 J/K then why do you have it listed as -217 J/K*mol? So is DS rxn = -217 J or 2*-217 J?
Same kind of question for DH.
DS = -217 J/(K)mol
DH = -1202 KJ -> -1202 x 10^3 J
does this make any sense?
To determine whether a reaction is spontaneous at a given temperature, we can use the equation for the change in the total entropy of the universe:
ΔSuniverse = ΔSsystem - ΔHsystem / T
In this case, we have:
ΔSsystem = -217 J/K*mol
ΔHsystem = -1202 kJ
T = 25°C = 298 K
Plugging in these values, we get:
ΔSuniverse = (-217 J/K*mol) - (-1202 kJ) / (298 K)
First, we need to convert -1202 kJ to J:
ΔHsystem = -1202 kJ × 1000 J/1 kJ = -1,202,000 J
Now we can continue with the calculation:
ΔSuniverse = -217 J/K*mol - (-1,202,000 J) / 298 K
ΔSuniverse = -217 J/K*mol + 4,037 K
ΔSuniverse = 3,820 J/K*mol
From the calculation, we can see that ΔSuniverse is positive, indicating that the reaction is spontaneous at 25°C/298 K. This means that the reaction will proceed on its own without the need for external assistance.
Your intuition was partially correct when considering the signs of ΔG and ΔH. The spontaneity of a reaction can generally be determined by the sign of ΔG, where a negative value for ΔG indicates a spontaneous reaction. However, in this case, we are using ΔSuniverse to determine spontaneity, which accounts for both the system and its surroundings. Since ΔSuniverse is positive, it implies an increase in the overall entropy of the universe, which drives the reaction to be spontaneous.
Thus, based on the positive value of ΔSuniverse, the reaction 2Mg(s) + O2(g) → 2MgO(s) is indeed spontaneous at 25°C/298 K.
To determine whether a reaction is spontaneous at a given temperature, you can use the Gibbs free energy equation:
ΔG = ΔH - TΔS
ΔG represents the Gibbs free energy change of the system, ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin.
In your case, ΔH is given as -1202 kJ (notice that the unit should be kilojoules, not kilojoules per mole), and ΔS is given as -217 J/K*mol. To use consistent units, convert ΔH to J by multiplying by 1000, so ΔH = -1202000 J.
Let's plug in the values into the equation:
ΔG = -1202000 J - T(-217 J/K*mol)
At 25°C, T is 298 K. Substituting this value:
ΔG = -1202000 J - (298 K)(-217 J/K*mol)
ΔG = -1202000 J + 64866 J
ΔG = -1137134 J
Since the value of ΔG is negative, the reaction is spontaneous at 25°C. A negative ΔG indicates that the reaction is energetically favorable and can occur without the need for an external energy source.
You mentioned that you obtained a positive value for ΔSuniverse. It's important to note that while ΔSuniverse can provide some insights into the spontaneity of a reaction, it is ΔG that is the definitive indicator. In this case, the negative ΔG suggests that the reaction will proceed spontaneously, even if ΔH is negative (indicating an endothermic reaction) and ΔS is positive (indicating an increase in disorder).