In: Chemistry
1. Suppose 50.00 mL of 2.0 × 10–4 M Fe(NO3)3 is added
to 50.00 mL of 2.0 ×10-6 M KIO3. Which of the following
statements is true? For Fe(IO3)3, Ksp = 1.0 ×
10–14.
A) A precipitate forms because Qc >
Ksp.
B) A precipitate forms because Qc <
Ksp.
C) No precipitate forms because Qc < Ksp.
D) No precipitate forms because Qc = Ksp.
E) No precipitate forms because Qc > Ksp.
2. For which of the following reactions is ∆S° > 0 at 25°C?
A) 2H2(g) + O2(g) → 2H2O(g)
B) 2ClBr(g) → Cl2(g) + Br2(g)
C) I2(g) → I2(s)
D) 2NO(g) + O2(g) → 2NO2(g)
E) NH4HS(s) → NH3(g) + H2S(g
3. What is E of the following cell reaction at 25°C? Cu(s) | Cu2+(0.017 M) || Ag(s), (Ag+ = 0.18M)
E°cell = 0.460 V.
A) 0.468V
B) 0.282 V
C) 0.460 V
D) 0.490 V
E) 0.479V
Q1 Suppose 50.00 mL of 2.0 × 10–4 M Fe(NO3)3 is added to 50.00 mL of 2.0 ×10-6 M KIO3. Which of the following statements is true? For Fe(IO3)3, Ksp = 1.0 × 10–14.
Solution :-
Using the given concentration values lets calculate the Qc
Qc equation is as follows
Qc=[Fe^3+][IO3^-]3
Lets put the values in the formula after adding the solution total volume will become 100 ml therefore the initial concentrations will halved.
[Fe^3+] = 2.0E-4 / 2 = 1.0E-4 M
[IO3^-] = 2.0E-6 / 2 = 1.0E-6 M
Qc = [1.0E-4 ] [1.0E-6]2
Qc = 1E-16
Therefore Qc is smaller than Kc means reaction will proceed in forward direction.
Therefore correct answer is
No precipitate forms because Qc < Ksp.
Q2 solution
For the reaction
NH4HS(s) → NH3(g) + H2S(g)
Delta S > 0at 25 c
Because here solid is changing to the gases.Gases have the greatest entropy than liquid and solid.
Therefore the reaction shown in the option E will have entropy greater than 0
Q3 solution :-
Cu(s) | Cu2+(0.017 M) || Ag(s), (Ag+ = 0.18M) E°cell = 0.460 V.
Formula to calculate the E of cell is as follows
E cell = Eo cell – (0.0592/n)logQ
Where log Q = [product ]/[reactant]
N=number of electrons transferred.
Lets put the values in the formula
E = 0.460 V – (0.0592/2)log [0.017/0.18]
E= 0.460 V – (-0.0303)
E= 0.490 V
Therefore correct answer is option ‘D’