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rate = (- ∆ [product] )/(∆ t)
2BF_3 + 3 Li_2 SO_3 → B_2 (SO_3 )_( 3) + 6 LIF
B_2 (SO_3 )_3 Will change at the faster rate.
Question 13
Initial ph of the hbr solution
HBr + KOH → H_2 O + KBr
molarity of HBr = 2.4 ×10^( -2) M
molarity of KOH = 1.4 × 1〖0 〗^(-2) M
pH of HBr = -log [ 2.4 ×10^( -2) M]
pH of HBr = 1.6197887580
Initial poh of the hbr solution
HBr + KOH → H_2 O + KBr
molarity of HBr = 2.4 ×10^( -2) M
molarity of KOH = 1.4 × 1〖0 〗^(-2) M
The relation of ph and poh
1 × 10^( -14 ) = [H^( +) ] [OH^- ]
[H^( +) ] = 1.6197887580
1 × 10^( -14 ) = 1.6197887580 [OH^- ]
( ( 1 × 10^( -14 ) ))/( 1.6197887580 ) = [OH^- ]
[OH^- ] = 6.17× 10^( -15)
pH + pOH = 14
pOH = 14 - 1.6197887580
pOH = 12.38021124
Volume of KOH for equivalence point
initial volume of the solution = 15 mL
mol base = mol acid = 15 ml × 2.4 ×10^( -2) M= 0.36 mmol
= 0.36 m mol × (m L)/( 1.4 ×10^( -2) m mol)
volume of KOH required to reach the equivalence point = 25.71 mL
Concentration after adding more acid
new volume added = 2 mL = 0.002 M
moles of KOH = 1.4 × 10^( -2)
initial volume = 15 mL
initial moles of KOH = 2.4 × 10^( 2) M
total final volume= 15 ml + 2 mL = 17 ml
= ( 17)/( 1000) = 0.017 M
= 0.002 - 1.4 × 10^( -2) = 0 .012
final concentration = ( 0.012)/( 0.017)
final concentration = 0.705882353 M
Final ph of the solution after adding 2ml of KOH
pH of KOH = -log [ H_( 3) O^( +) ]
pH of KOH = -log [ 0.705882353 ]
pH of KOH = 0.151267675
Question 14
〖H 〗_3 (I) +Br_( 2) (I) + H_( 3) O^( + )→ CH_( 3) COCH_( 2) Br (aq) + H_( 3) O^( +) (aq) + Br^( -) (aq)
If the rate law is given by the equation: rate = k [CH_( 3) COCH_( 3) ]^( X) [ B〖r 〗_(2 ) ]^( Y ) [H_3 O]^( Z)What are the reaction orders x, y and z?
Trial [CH_( 3) COCH_( 3) ] ( M ) Br_( 2) [M] H_( 3) O^( +) (M) Initial rate (M/s)
1 0.30 0.050 0.050 5.7 × 10^( -5)
2 0.30 0.100 0.050 5.7 × 10^( -5)
3 0.30 0.050 0.200 1.2 × 10^( -4)
4 0.40 0.050 0.200 3.1 × 10^( -4)
5 0.40 0.050 0.050 7.6 × 10^( -5)
Trial 1, 2
rate = k [CH_( 3) COCH_( 3) ]^( X) [ B〖r 〗_(2 ) ]^( Y ) [H_3 O]^( Z)
5.7 × 10^( -5) = k [0.30]^( X) [ 0.050]^( Y ) [0.050]^( Z)
5.7 × 10^( -5) = k [0.30]^( X) [ 0.100]^( Y ) [0.050]^( Z)
Dividing the second equation by first gives
1 = [( 0.100)/( 0.050)]^( y)
1 = [2.00]^( y)
y = 0
Trial 3, 4
rate = k [CH_( 3) COCH_( 3) ]^( X) [ B〖r 〗_(2 ) ]^( Y ) [H_3 O]^( Z)
1.2 × 10^( -4) = k [0.30]^( X) [ 0.050]^( Y ) [0.200]^( Z)
3.1 × 10^( -4) = k [0.40]^( X) [ 0.050]^( Y ) [0.200]^( Z)
Dividing the second equation by first gives
2.58 = [( 0.400)/( 0.300)]^( X)
2.58 = [1.33]^( X)
x = 3.32
Trial 4, 5
rate = k [CH_( 3) COCH_( 3) ]^( X) [ B〖r 〗_(2 ) ]^( Y ) [H_3 O]^( Z)
3.1 × 10^( -4) = k [0.40]^( X) [ 0.050]^( Y ) [0.020]^( Z)
7.6 × 10^( -5) = k [0.40]^( X) [ 0.050]^( Y ) [0.050]^( Z)
Dividing the second equation by first gives
0.2451 = [( 0.050)/( 0.020)]^( Z)
0.24516 = [2.5]^( Z)
z = 1.53
Rate becomes
rate = k [CH_( 3) COCH_( 3) ]^( 3.32) [ B〖r 〗_(2 ) ]^( 0 ) [H_3 O]^( 1.53)
Question 15
In the first trail the data used to determine the rate law can be used here,
rate = k [CH_( 3) COCH_( 3) ]^( X) [ B〖r 〗_(2 ) ]^( Y ) [H_3 O]^( Z)
[ B〖r 〗_(2 ) ]^( 0 ) = 1
5.7 × 10^( -5) = k [0.30]^( X) [ 0.050]^( Y ) [0.050]^( Z) = k [CH_( 3) COCH_( 3) ]^( 3.32) [H_3 O]^( 1.53)
In first trial
CH_( 3) COCH_( 3) = 0.3
H_3 O = 0.05
5.7 × 10^( -5) = k [0.3]^( 3.32) [0.05]^( 1.53)
k = 0.303674
Question 16
Hydronium to increase the reaction rate
The increase of hydronium concentration in the solution increases the amount of hydroxide ion and the solution produce several ions. The increasing concentration of hydronium and hydroxide ion shift the equilibrium that the reaction rate increase. The shift in the concentration increase and decrease the rates of Bronsted acid catalyzed reactions.
Question 17
Reduction equation of galvanic cell based on Au and Ag
The working of galvanic cell is based on the movement of Au and Ag ions in the solution. The process is based on the redox and oxidation process at the anode and cathode. The reduction reaction equations are listed below,
Au^( 3+) + 3e^( -) → Au^( 0) 〖E 〗^0 = + 1.50
Ag^( + ) + 1e^( - ) → Ag^( 0) E^( 0) = +0.80
Question 18
Equation for the oxidation of half cell reaction
In the galvanic cell the redox and oxidation reactions are between gold and silver. the oxidation reaction takes, and redox reactions are listed below,
Au^( 3+ ) + 3Ag → 3Ag^( +) + Au
Ag^( + ) + 1e^( - ) → Ag^( 0)
Au^( 3+) + 3e^( -) → Au^( 0)
Question 19
Electrodes in galvanic cell
In the galvanic cells, the salt bridge is used to provide connection between two regions of the cell. The Ag ions moves towards the cathode half cell while Au moves towards the anode of the cell. The movement of ions depends on the nature of the ions.
Question 20
Calculation of E0 cell reaction
Calculating the standard cell potential,
Half reaction:
Au^( + 3) + 3〖 e〗^( -) → Au +1.50
Ag^( + ) + 1e^( - ) → Ag^( 0) E^( 0) = +0.80
E_( Au)^( 0 ) = 1.50
E_( Ag)^( 0) = 0.80
E_( cell)^( 0) = E_( red)^( 0) - E_( oxid)^( 0)
E_( cell)^( 0) = 0.80 - 1.50
E_( cell)^( 0) = -0.7 V
Question 21
Oxidizing agents
Au^( 3+ ) + 3Ag → 3Ag^( +) + Au
Ag^( +) = oxidizing agent that cause Au to get reduced
Reducing agents
Au^( +) = reducing agent that cause Ag to get reduced
It loses electrons in the redox reaction.
Question 22
The overall reaction
A + E + 2B → C + D
Molecularity of step 1
Molecularity is simply sum of numbers of the molecules of reactants that are constituents of reaction. The molecularity is sum of number of molecules and in first equation it will be 2.
2B → C + E
Molecularity of step 2
In the second step, the molecularity is sum of two components in the reaction and it becomes 2.
Question 23
The reaction equation of the reaction that consist of two steps is the cumulative reaction equation and it is listed below,
rate= k [B]^( 2)
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