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Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
The ionic compound of FePO4 is the iron III phosphate. Or you can call it ferric phosphate or ferric orthophosphate.
Just for general informations, It's normally used in steel and metal manufacturing processes, in organic farming, and many other uses.
Hope this Helps :)
Answer: 600 kJ
-
Explanation:
C₃H₈ (g) + 5 O₂ (g) =============== 3 CO₂ (g) + 4 H₂O (l)
Δ⁰Hf kJ/mol -104 0 -393.5 -285.8
Δ⁰Hcomb C₃H₈ = 3(-393.5) + 4 (-285.80) - (-104) kJ/mol
Δ⁰Hcomb = 2219.70 kJ/mol
n= m /MW MW c₃H₈ = 44.1 g/mol
n= 12 g/44.1 g/mol = 0.27 mol
then for 12 g the heat released will be
0.27 mol x 2219.70 kJ/mol = 600 KJ
