Consider 1.5 L of air in a patents lungs at 37.0 C and 1.00 atm pressure. What is the volume would this air occupy if it were at
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Answer:
Explanation:
Relation between ΔG₀ and K ( equilibrium constant ) is as follows .
lnK = - ΔG₀ / RT
The value of R and T are same for all reactions .
So higher the value of negative ΔG₀ , higher will be the value of K .
Mg(s) + N₂0(g) → MgO(s) + N₂(g)
has the ΔG₀ value of -673 kJ which is highest negative value . So this reaction will have highest value of equilibrium constant K .
The volume of H₂O = 5 L
<h3>Further explanation</h3>Given
5L of H₂ and 3L O₂
Reaction
2H₂ (g) + O₂(g) ⇒2H₂O(g)
Required
The volume of H₂O
Solution
Avogadro's hypothesis:
<em>In the same T,P and V, the gas contains the same number of molecules </em>
So the ratio of gas volume will be equal to the ratio of gas moles
mol H₂ = 5, mol O₂ = 3
- Find limiting reactants
From equation, mol ratio H₂ : O₂ = 2 : 1, so :
- Find volume H₂O
mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L