Hellium argon neon xenon krypton randon oxygen fluorine chlorine bromine
We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula:
Kc = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 = 0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07
Answer:
B = (2.953 × 10⁻⁹⁵) N.m⁹
Explanation:
At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.
That is,
Fa + Fr = 0
Fa = - Fr
Fa = (|q₁q₂|)/(4πε₀r²)
Fr = -B/(r^n) but n = 9
Fr = -B/r⁹
(|q₁q₂|)/(4πε₀r²) = (B/r⁹)
|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C
(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²
r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m
(k|q₁q₂|)/(r²) = (B/r⁹)
(k × |q₁q₂|) = (B/r⁷)
B = (k × |q₁q₂| × r⁷)
B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]
B = (2.953 × 10⁻⁹⁵) N.m⁹