11th Class Chemistry Chapter 8 Chemical Equilibrium Short Question Answers below
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H2 + l2⇋ 2Hl Kc = [Hl]2/[H2][l2]
If the number of mole of reactants and products are unequal then Kc has net units
PCl3 ⇋ PCl3 + Cl2 Kc = [cons]+1
N2 + 3H2 ⇋ 2NH3 Kc = [conc]-2
When we have such reversible reaction, in which the total number of moles of reactants and products are equal, then Kp = Kx = Kn. In other words, Δn = 0, where Δn is the deference of number of moles of reactants and products.
In the following equation Kp and Kc are related by
Kp = Kc (RT)Δn
H2 + l2 ⇋ 2Hl Kc = 4x2/(a-x)(b-x)
(i). Q < Kc. The reaction goes to the forward direction.
(ii). Q < Kc. The reaction goes to the backward direction.
(iii). Q < Kc
N2 + 3H2 ⇋ 2NH3
In 2O2 + O2 ⇋ 2SO3, the same principle of pressure is applicable.
-log [H+] = +pH
log [H+] = -pH = -10
[H+] = 10 -10
Since [H+] [OH] = 10-10 So [OH–] = 10-4
So pH + pOH 14= pKw
When temperature of the aqueous solution is above 250C then
pH + pOH < 14
pH + pOH = pKw
[OH–] = 2* 10-4 mol dm-3 pH -log2 * 10-4
According to protonic concept, acids are those species which are proton donors or have a tendency to donate a porton.
CH3COOH + H2O ⇋ CH3COOH– + H3O+
weak acid base conjugate conjugate
CH3COOH has a little tendency to break the oxygen hydrogen bond and to act as an acid. It means that CH3COOH– ion should have a strong tendency to react with a proton. So CH3COOH– should be a strong base. HCl is a very strong acid and so Cl– ion should have a least tendency to accept proton to act as a base.
NaCl(s) ⇋ Na+ (aq) + Cl–(aq)
If HCl gas in passed in saturated solution of NaCl, the Cl– ions are generated in express in the solution
HCl(aq) ⇋ H+(aq) + Cl–(aq)
Due to excess or Cl–, the ionization of NaCl is suppressed and NaCl settles down in the form of precipitate.
Na+(aq)+ Cl–(aq) ⇋ NaCl(s) (ionization is suppressed)
Moreover the use of HCl in ll- group basic radicals is for common ion effect.
(i). Weak acid + salt with a strong base.
(ii). Weak base + salt with a strong acid.
pH = pKa + log [salt]/[Acid] and POH = pKb + log [Salt]/[Base]
pH = pKa + log [salt]/[acidΔ
The pKa of the acid, which is being used for the preparation of this buffer solution is a constant quantity at a given temperature. By adjusting the ratio of salt and acid concentration, we can prepare a solution having pH of own choice.
pH = pKa + log [Salt]/[acid]
According to this equation, when the [salt] is greater than [acid], then the long term becomes bigger and the pH becomes higher. In other words, the solution obtained is of lower acid strength and of higher pH.
CH3COOH ⇋ CH3COO–(aq) + H+(aq)
CH3COONa⇋ CH3COO–(aq) + Na+(aq)
When a few drops of an acid , say HCl are added I this solution, the H+ ions provided by HCl are taken up by CH3COO– (mostly obtained from CH3COONa), so incoming protons are consumed and pH is retained. When a few drops of a base say NaOH is added from outside, then the protons already present in the solution are consumed. To compensate those protons, there happens a further dissociation of CH3COOh and pH is retained.
pKa + pKb = 14
Take the log of equation
log Ka + Kb = log Kw
Log Ka + log Kb = log Kw
Multiply it with negative sign
-log Ka – log Kb = -logKw
pKa + pKb = pKw
Since pKw= 14, of 250C so the pKa and pKb of the conjugate acid base pair has a very simple relationship with each other.
HCl ⇋ H+ + Cl–
10-4 mole dm-3 ⇋ 0+0 t = 0
0 ⇋ 10-4 mol dm-3 t= Eq]
So, [H+] = 10-4
-log[H+] = pH = 4
NH4OH ⇋ NH4– + OH–
NH4Cl ⇋ NH4+ + Cl
Since OH– ions are produced in the solution pH will be more than 7, and solution will be basic.
(ii). If [M+][X–] = Ksp, solution is saturated.
(iii). If [M+][X–] > Ksp , solution is momentarily supersaturated.
N2(g) + 3H2(g) ⇋ 2NH3(g)
Kc = [NH3][N2][H2]3 = (mol dm-3)-2
(mol dm-3)(mol dm-3)3 (mol dm-3)-2
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