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 potential energy of the reactants is 200J.
From the energy diagram, the energy of the product formed is 350J; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.<span />
Since there is so little information given, I will assume that we are at STP and i can use the conversion factor at STP--->> 22.4 Liters= 1 mol of gas
before we use this conversion, we need to convert the grams to moles using the molar mass of the molecule.
molar mass of Cl₂= 35.5 x 2= 71.0 g/ mol
177.3 g (1 mol/ 71.0 g)= 2.50 mol Cl₂
then we use the conversion to get the volume
2.50 mol Cl₂ (22.4 Liters/ 1 mol)= 55.9 Liters
I could not break unless you hit it with a sledge hammer
Answer:
on the surface of the cathode
