When the oxide of generic metal m is heated at 25.0 °c, only a negligible amount of m is produced. o2m(s ----> m(s o2(g delta
1 answer:
(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸
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Answer:
The reaction will be spontaneous
Explanation:
To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:
<u>If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.</u>
Calculating the :
Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.
In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in:
If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.