Answer:
Part 1: - 1.091 x 10⁴ J/mol.
Part 2: - 1.137 x 10⁴ J/mol.
Explanation:
Part 1: At standard conditions:
At standard conditions Kp= 81.9.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.
Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.
For the reaction:
I₂(g) + Cl₂(g) ⇌ 2ICl(g).
Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.
First we determine the
moles CaCl2 present:
525g / (110.9g/mole) =
4.73 moles CaCl2 present
Based on stoichiometry,
there are 2 moles of Cl for every mole of CaCl2:<span>
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl </span>
Get the mass:<span>
<span>9.47moles Cl x 35.45g/mole = 335.64 g Cl</span></span>
Answer is 14.5 g L⁻¹.
<em>Explanation;</em>
Here, the question says reduce the units as one.
The presented units are g/L. To reduce the units to one, what we can do is take L to the upper side.
This can be done according to the rules of indices;
1 / aˣ = a⁻ˣ
Like that, we can write 1 / L as L⁻¹.
Hence, the reduced unit is g L⁻¹.
But remember to keep a space between when writing two different units.
Actually, this is an unit for density.
Intermolecular force for solids is high. Whereas low in gases. The smell of agarbatti spreads immediately because the molecules of air diffuses very fastly.
