<span>H2S has 2 bonds which means that it has two shared electron pairs. It has 2 lone pairs on electrons. It's the only one where the 2 are the same.</span>
The Balanced Chemical Equation is as follow;
4 KO₂ + 2 CO₂ → 2 K₂CO₃ + 3 O₂
First find out the Limiting Reagent,
According to equation,
284 g (4 moles) KO₂ reacted with = 44.8 L (2 moles) of CO₂
So,
27.9 g of KO₂ will react with = X L of CO₂
Solving for X,
X = (44.8 L × 27.9 g) ÷ 284 g
X = 4.40 L of CO₂
Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,
According to eq.
284 g (4 moles) KO₂ formed = 138.2 g of K₂CO₃
So,
27.9 g of KO₂ will form = X g of K₂CO₃
Solving for X,
X = (138.2 g × 27.9 g) ÷ 284 g
X = 13.57 g of K₂CO₃
So, 13.57 g of K₂CO₃ formed is the theoretical yield.
%age Yield = 13.57 / 21.8 × 100
%age Yield = 62.24 %
There are several units for expressing energy, Most common are joules and kilocalories.
The conversion factor of joule to kilocalories is
1 kilo calorie = 4184 J
∴ x kilo calorie = 256 J
x =
kilo calories
Thus, 256 J = 0.0611 kilocalories
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
