Answer:
(D) K = 4.5 x 10⁵. The reaction favors the formation of products
Explanation:
HCOOH + CN⁻ ⇆ HCOO⁻ + HCN
K = [HCOO⁻] [ HCN ] / [ HCOOH] [ CN⁻]
HCOOH ⇄ H ⁺ + COO⁻
K₁ = [ H⁺] [ COO⁻ ] / [HCOOH ]
HCN ⇆ H⁺ + CN⁻
K₂ = [ H⁺] [ CN⁻] / [ HCN ]
K₁ / K₂
= [ H⁺] [ COO⁻ ] / [HCOOH ] X [ HCN ] / [ H⁺] [ CN⁻]
= [ COO⁻ ][ HCN ] / [HCOOH ] [ CN⁻]
= K
K = K₁ / K₂
= 1.8 x 10⁻⁴ / 4 x 10⁻¹⁰
= 4.5 x 10⁵
So equilibrium constant of the reaction
HCOOH + CN⁻ ⇆ HCOO⁻ + HCN
is very high . Hence reaction favours the formation of product.
option (D) is correct.
In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for electrons. II. the oxidized form has a lower affinity for electrons. III. the reduced form has a higher affinity for electrons. IV. the greater the tendency for the oxidized form to accept electrons.
Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.13 m FeCl3 A. Highest boiling point 2. 0.19 m Mg(CH3COO)2 B. Second highest boiling point 3. 0.30 m KI C. Third highest boiling point 4. 0.53 m Glucose(nonelectrolyte) D. Lowest boiling point An error has been detected in your answer. Check for typos,
Answer:0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point-C
0.13 m FeCl3---- Lowest boiling point-D
Explanation:
Using the boilng point elevation formula
ΔTb=m* kb *i
where m= molality
kb= elevated boiling point constant( here kb values will be same for all soluton)
i= vant hoff factor = number of ions present in a solution
Using the number of ions and molarity present in a solution as a collagative property, since kb is constant, we can determine which of the species has the highest boiling point.
1.) 0.13 m FeCl3= Fe³⁻ + Cl⁻
i=4
ΔTb=m* kb* i= molarity x number of ionsx Kb= 0.13 x 4= 0.52kb
2) 0.19 m Mg(CH3COO)2 = Mg²⁺ + CH₃COO⁻
i= 3
ΔTb=m* kb* i= molarity x number of ions= 0.19 x 3= 0.57kb
3. 0.30 m KI = K⁺ + I⁻
i= 2
ΔTb=m *kb *= imolarity x number of ions xKb= 0.30x 2= 0.60kb
4. 0.53 m Glucose(nonelectrolyte) =
i= 1 for nonelectroytes
ΔTb=m* kb* i = molarity x number of ionsx Kb= 0.53 x 1= 0.53Kb
therefore,
0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point
0.13 m FeCl3---- Lowest boiling point