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
<span>For equation A + 3B + 2C ---> 2D,
1 mole of A will produce 2 moles of D
3 moles of B will produce 2 moles of D, so 1 mole of B will produce 2/3 moles of D
2 moles of C will produce 2 moles of D, so 1 mole of C will produce 1 mole of D
If only 1 mole of B is present, only 2/3 moles of D can be produced. This is regardless of the number of moles of A and C. B is the limiting reactant and the maximum number of moles of D expected is 2/3.</span>
A valid Lewis structure of IF3 cannot be drawn without violating the octet rule.
Answer: IF3 (Iodine Trifluoride)
This is because, I (Iodine) and F (Fluorine) both have odd number of valence electrons (7) which also means that there are too many valence electrons in the valence shell.
Answer:
The Relative Formula Mass of Fe(NO₃)₂ is 179.8524 grams
Explanation:
The Relative Formula Mass is the mass of one mole of a compound expressed in grams, obtained by adding together the Relative Atomic Masses, RAM, of the elements which makes the compound
The Relative Formula Mass of a compound is the same as its Relative Molecular Mass
The relative formula mass of Fe(NO₃)₂ is given as follows;
The relative atomic mass of Fe = 55.845 amu
The relative atomic mass of nitrogen, N = 14.0067 amu
The relative atomic mass of oxygen, O = 15.999 amu
Therefore, we have;
The formula mass of Fe(NO₃)₂ = (55.845 + 2×(14.0067 + 3×15.999)) amu = 179.8524 amu
The Relative Formula Mass of Fe(NO₃)₂ = 179.8524 grams.
