The answer would be 2 in ths case, bonding has to do with the electron orbitals.
Answer:
9.2
Explanation:
Let's do an equilibrium chart of this reaction:
2NO(g) + O₂(g) ⇄ 2NO₂(g)
4.9 atm 5.1 atm 0 Initial
-2x -x +2x Reacts (stoichiometry is 2:1:2)
4.9-2x 5.1-x 2x Equilibrium
The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:
y = 2x/(4.9 - 2x + 5.1 -x + 2x)
0.52 = 2x/(10 - x)
2x = 5.2 -0.52x
2.52x = 5.2
x = 2.06 atm
Thus, the partial pressure at equilibrium are:
pNO = 4.9 -2*2.06 = 0.78 atm
pO₂ = 5.1 - 2.06 = 3.04 atm
pNO₂ = 2*2.06 = 4.12 atm
Thus, the pressure equilibrium constant Kp is:
Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]
Kp = [(4.12)²]/[(0.78)²*3.04]
Kp = [16.9744]/[1.849536]
Kp = 9.2
D: is the anwser that your looking for
Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
