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1 | 3205-3208 | 6
Example 3 6
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
76
Chemistry
The half-life of a reaction is the time in which the concentration of a
reactant is reduced to one half of its initial concentration It is
represented as t1/2 For a zero order reaction, rate constant is given by equation 3 |
1 | 3206-3209 | 6
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
76
Chemistry
The half-life of a reaction is the time in which the concentration of a
reactant is reduced to one half of its initial concentration It is
represented as t1/2 For a zero order reaction, rate constant is given by equation 3 7 |
1 | 3207-3210 | It is
represented as t1/2 For a zero order reaction, rate constant is given by equation 3 7 [
]
[ ]
R0
R
k
t
−
=
[
]
[
]0
1/2
1 R
At
,
R
2
t
=t
=
The rate constant at t1/2 becomes
[
]
[
]
0
0
1/2
1/2
R
R
k
−t
=
[ ]0
1/2
R
2
t
k
=
It is clear that t1/2 for a zero order reaction is directly proportional
to the initial concentration of the reactants and inversely proportional
to the rate constant |
1 | 3208-3211 | For a zero order reaction, rate constant is given by equation 3 7 [
]
[ ]
R0
R
k
t
−
=
[
]
[
]0
1/2
1 R
At
,
R
2
t
=t
=
The rate constant at t1/2 becomes
[
]
[
]
0
0
1/2
1/2
R
R
k
−t
=
[ ]0
1/2
R
2
t
k
=
It is clear that t1/2 for a zero order reaction is directly proportional
to the initial concentration of the reactants and inversely proportional
to the rate constant For the first order reaction,
[
]
[
]
0
2 |
1 | 3209-3212 | 7 [
]
[ ]
R0
R
k
t
−
=
[
]
[
]0
1/2
1 R
At
,
R
2
t
=t
=
The rate constant at t1/2 becomes
[
]
[
]
0
0
1/2
1/2
R
R
k
−t
=
[ ]0
1/2
R
2
t
k
=
It is clear that t1/2 for a zero order reaction is directly proportional
to the initial concentration of the reactants and inversely proportional
to the rate constant For the first order reaction,
[
]
[
]
0
2 303
R
log
R
k
t
=
(3 |
1 | 3210-3213 | [
]
[ ]
R0
R
k
t
−
=
[
]
[
]0
1/2
1 R
At
,
R
2
t
=t
=
The rate constant at t1/2 becomes
[
]
[
]
0
0
1/2
1/2
R
R
k
−t
=
[ ]0
1/2
R
2
t
k
=
It is clear that t1/2 for a zero order reaction is directly proportional
to the initial concentration of the reactants and inversely proportional
to the rate constant For the first order reaction,
[
]
[
]
0
2 303
R
log
R
k
t
=
(3 15)
at t1/2 [
]
[ ]0
R
R
2
=
(3 |
1 | 3211-3214 | For the first order reaction,
[
]
[
]
0
2 303
R
log
R
k
t
=
(3 15)
at t1/2 [
]
[ ]0
R
R
2
=
(3 16)
So, the above equation becomes
[
]
[
]
0
1/2
2 |
1 | 3212-3215 | 303
R
log
R
k
t
=
(3 15)
at t1/2 [
]
[ ]0
R
R
2
=
(3 16)
So, the above equation becomes
[
]
[
]
0
1/2
2 303
R
log
R/2
k
t
=
or
1/2
2 |
1 | 3213-3216 | 15)
at t1/2 [
]
[ ]0
R
R
2
=
(3 16)
So, the above equation becomes
[
]
[
]
0
1/2
2 303
R
log
R/2
k
t
=
or
1/2
2 303 log 2
t
k
1/2
2 |
1 | 3214-3217 | 16)
So, the above equation becomes
[
]
[
]
0
1/2
2 303
R
log
R/2
k
t
=
or
1/2
2 303 log 2
t
k
1/2
2 303
0 |
1 | 3215-3218 | 303
R
log
R/2
k
t
=
or
1/2
2 303 log 2
t
k
1/2
2 303
0 301
t
k
=
×
1/2
0 |
1 | 3216-3219 | 303 log 2
t
k
1/2
2 303
0 301
t
k
=
×
1/2
0 693
t
k
=
(3 |
1 | 3217-3220 | 303
0 301
t
k
=
×
1/2
0 693
t
k
=
(3 17)
2
pN O5
= 1 |
1 | 3218-3221 | 301
t
k
=
×
1/2
0 693
t
k
=
(3 17)
2
pN O5
= 1 5 – 2 × 0 |
1 | 3219-3222 | 693
t
k
=
(3 17)
2
pN O5
= 1 5 – 2 × 0 512 = 0 |
1 | 3220-3223 | 17)
2
pN O5
= 1 5 – 2 × 0 512 = 0 476 atm
Using equation (3 |
1 | 3221-3224 | 5 – 2 × 0 512 = 0 476 atm
Using equation (3 16)
i
A
4
1
0 |
1 | 3222-3225 | 512 = 0 476 atm
Using equation (3 16)
i
A
4
1
0 5 atm
2 |
1 | 3223-3226 | 476 atm
Using equation (3 16)
i
A
4
1
0 5 atm
2 303
2 |
1 | 3224-3227 | 16)
i
A
4
1
0 5 atm
2 303
2 303
log
100slog
0 |
1 | 3225-3228 | 5 atm
2 303
2 303
log
100slog
0 476 atm
2 |
1 | 3226-3229 | 303
2 303
log
100slog
0 476 atm
2 303
0 |
1 | 3227-3230 | 303
log
100slog
0 476 atm
2 303
0 0216
4 |
1 | 3228-3231 | 476 atm
2 303
0 0216
4 98
10
s
100s
p
k
t
p
3 |
1 | 3229-3232 | 303
0 0216
4 98
10
s
100s
p
k
t
p
3 3 |
1 | 3230-3233 | 0216
4 98
10
s
100s
p
k
t
p
3 3 3 Half-Life of
a Reaction
Rationalised 2023-24
77
Chemical Kinetics
A first order reaction is found to have a rate constant, k = 5 |
1 | 3231-3234 | 98
10
s
100s
p
k
t
p
3 3 3 Half-Life of
a Reaction
Rationalised 2023-24
77
Chemical Kinetics
A first order reaction is found to have a rate constant, k = 5 5 × 10-14 s-1 |
1 | 3232-3235 | 3 3 Half-Life of
a Reaction
Rationalised 2023-24
77
Chemical Kinetics
A first order reaction is found to have a rate constant, k = 5 5 × 10-14 s-1 Find the half-life of the reaction |
1 | 3233-3236 | 3 Half-Life of
a Reaction
Rationalised 2023-24
77
Chemical Kinetics
A first order reaction is found to have a rate constant, k = 5 5 × 10-14 s-1 Find the half-life of the reaction Half-life for a first order reaction is
t1/2
=
0 |
1 | 3234-3237 | 5 × 10-14 s-1 Find the half-life of the reaction Half-life for a first order reaction is
t1/2
=
0 693
k
t1/2
=
–14
–1
0 |
1 | 3235-3238 | Find the half-life of the reaction Half-life for a first order reaction is
t1/2
=
0 693
k
t1/2
=
–14
–1
0 693
5 |
1 | 3236-3239 | Half-life for a first order reaction is
t1/2
=
0 693
k
t1/2
=
–14
–1
0 693
5 5×10
s
= 1 |
1 | 3237-3240 | 693
k
t1/2
=
–14
–1
0 693
5 5×10
s
= 1 26 × 1013s
Show that in a first order reaction, time required for completion of
99 |
1 | 3238-3241 | 693
5 5×10
s
= 1 26 × 1013s
Show that in a first order reaction, time required for completion of
99 9% is 10 times of half-life (t1/2) of the reaction |
1 | 3239-3242 | 5×10
s
= 1 26 × 1013s
Show that in a first order reaction, time required for completion of
99 9% is 10 times of half-life (t1/2) of the reaction When reaction is completed 99 |
1 | 3240-3243 | 26 × 1013s
Show that in a first order reaction, time required for completion of
99 9% is 10 times of half-life (t1/2) of the reaction When reaction is completed 99 9%, [R]n = [R]0 – 0 |
1 | 3241-3244 | 9% is 10 times of half-life (t1/2) of the reaction When reaction is completed 99 9%, [R]n = [R]0 – 0 999[R]0
k
=
0
2 |
1 | 3242-3245 | When reaction is completed 99 9%, [R]n = [R]0 – 0 999[R]0
k
=
0
2 303
R
log
R
t
=
0
0
0
2 |
1 | 3243-3246 | 9%, [R]n = [R]0 – 0 999[R]0
k
=
0
2 303
R
log
R
t
=
0
0
0
2 303
R
log
0 |
1 | 3244-3247 | 999[R]0
k
=
0
2 303
R
log
R
t
=
0
0
0
2 303
R
log
0 999
R
R
t
=
2 |
1 | 3245-3248 | 303
R
log
R
t
=
0
0
0
2 303
R
log
0 999
R
R
t
=
2 303 log103
t
t
= 6 |
1 | 3246-3249 | 303
R
log
0 999
R
R
t
=
2 303 log103
t
t
= 6 909/k
For half-life of the reaction
t1/2
= 0 |
1 | 3247-3250 | 999
R
R
t
=
2 303 log103
t
t
= 6 909/k
For half-life of the reaction
t1/2
= 0 693/k
1/2
t
t
= 6 |
1 | 3248-3251 | 303 log103
t
t
= 6 909/k
For half-life of the reaction
t1/2
= 0 693/k
1/2
t
t
= 6 909
10
0 |
1 | 3249-3252 | 909/k
For half-life of the reaction
t1/2
= 0 693/k
1/2
t
t
= 6 909
10
0 693
k
k
It can be seen that for a first order reaction, half-life period is
constant, i |
1 | 3250-3253 | 693/k
1/2
t
t
= 6 909
10
0 693
k
k
It can be seen that for a first order reaction, half-life period is
constant, i e |
1 | 3251-3254 | 909
10
0 693
k
k
It can be seen that for a first order reaction, half-life period is
constant, i e , it is independent of initial concentration of the reacting
species |
1 | 3252-3255 | 693
k
k
It can be seen that for a first order reaction, half-life period is
constant, i e , it is independent of initial concentration of the reacting
species The half-life of a first order equation is readily calculated from
the rate constant and vice versa |
1 | 3253-3256 | e , it is independent of initial concentration of the reacting
species The half-life of a first order equation is readily calculated from
the rate constant and vice versa For zero order reaction t1/2 µµµµµ [R]0 |
1 | 3254-3257 | , it is independent of initial concentration of the reacting
species The half-life of a first order equation is readily calculated from
the rate constant and vice versa For zero order reaction t1/2 µµµµµ [R]0 For first order reaction
t1/2 is independent of [R]0 |
1 | 3255-3258 | The half-life of a first order equation is readily calculated from
the rate constant and vice versa For zero order reaction t1/2 µµµµµ [R]0 For first order reaction
t1/2 is independent of [R]0 Example 3 |
1 | 3256-3259 | For zero order reaction t1/2 µµµµµ [R]0 For first order reaction
t1/2 is independent of [R]0 Example 3 7
Example 3 |
1 | 3257-3260 | For first order reaction
t1/2 is independent of [R]0 Example 3 7
Example 3 7
Example 3 |
1 | 3258-3261 | Example 3 7
Example 3 7
Example 3 7
Example 3 |
1 | 3259-3262 | 7
Example 3 7
Example 3 7
Example 3 7
Example 3 |
1 | 3260-3263 | 7
Example 3 7
Example 3 7
Example 3 7
Solution
Solution
Solution
Solution
Solution
Example 3 |
1 | 3261-3264 | 7
Example 3 7
Example 3 7
Solution
Solution
Solution
Solution
Solution
Example 3 8
Example 3 |
1 | 3262-3265 | 7
Example 3 7
Solution
Solution
Solution
Solution
Solution
Example 3 8
Example 3 8
Example 3 |
1 | 3263-3266 | 7
Solution
Solution
Solution
Solution
Solution
Example 3 8
Example 3 8
Example 3 8
Example 3 |
1 | 3264-3267 | 8
Example 3 8
Example 3 8
Example 3 8
Example 3 |
1 | 3265-3268 | 8
Example 3 8
Example 3 8
Example 3 8
Solution
Solution
Solution
Solution
Solution
Table 3 |
1 | 3266-3269 | 8
Example 3 8
Example 3 8
Solution
Solution
Solution
Solution
Solution
Table 3 4 summarises the mathematical features of integrated laws of
zero and first order reactions |
1 | 3267-3270 | 8
Example 3 8
Solution
Solution
Solution
Solution
Solution
Table 3 4 summarises the mathematical features of integrated laws of
zero and first order reactions Table 3 |
1 | 3268-3271 | 8
Solution
Solution
Solution
Solution
Solution
Table 3 4 summarises the mathematical features of integrated laws of
zero and first order reactions Table 3 4: Integrated Rate Laws for the Reactions of Zero and First Order
Order
Reaction
Differential
Integrated
Straight
Half-
Units of k
type
rate law
rate law
line plot
life
0
R® P
d[R]/dt = -k
kt = [R]0-[R]
[R] vs t
[R]0/2k
conc time-1
or mol L–1s–1
1
R® P
d[R]/dt = -k[R]
[R] = [R]0e-kt
ln[R] vs t
ln 2/k
time-1 or s–1
or kt =
ln{[R]0/[R]}
Rationalised 2023-24
78
Chemistry
Most of the chemical reactions are accelerated by increase in temperature |
1 | 3269-3272 | 4 summarises the mathematical features of integrated laws of
zero and first order reactions Table 3 4: Integrated Rate Laws for the Reactions of Zero and First Order
Order
Reaction
Differential
Integrated
Straight
Half-
Units of k
type
rate law
rate law
line plot
life
0
R® P
d[R]/dt = -k
kt = [R]0-[R]
[R] vs t
[R]0/2k
conc time-1
or mol L–1s–1
1
R® P
d[R]/dt = -k[R]
[R] = [R]0e-kt
ln[R] vs t
ln 2/k
time-1 or s–1
or kt =
ln{[R]0/[R]}
Rationalised 2023-24
78
Chemistry
Most of the chemical reactions are accelerated by increase in temperature For example, in decomposition of N2O5, the time taken for half of the
original amount of material to decompose is 12 min at 50oC, 5 h at
25oC and 10 days at 0oC |
1 | 3270-3273 | Table 3 4: Integrated Rate Laws for the Reactions of Zero and First Order
Order
Reaction
Differential
Integrated
Straight
Half-
Units of k
type
rate law
rate law
line plot
life
0
R® P
d[R]/dt = -k
kt = [R]0-[R]
[R] vs t
[R]0/2k
conc time-1
or mol L–1s–1
1
R® P
d[R]/dt = -k[R]
[R] = [R]0e-kt
ln[R] vs t
ln 2/k
time-1 or s–1
or kt =
ln{[R]0/[R]}
Rationalised 2023-24
78
Chemistry
Most of the chemical reactions are accelerated by increase in temperature For example, in decomposition of N2O5, the time taken for half of the
original amount of material to decompose is 12 min at 50oC, 5 h at
25oC and 10 days at 0oC You also know that in a mixture of potassium
permanganate (KMnO4) and oxalic acid (H2C2O4), potassium
permanganate gets decolourised faster at a higher temperature than
that at a lower temperature |
1 | 3271-3274 | 4: Integrated Rate Laws for the Reactions of Zero and First Order
Order
Reaction
Differential
Integrated
Straight
Half-
Units of k
type
rate law
rate law
line plot
life
0
R® P
d[R]/dt = -k
kt = [R]0-[R]
[R] vs t
[R]0/2k
conc time-1
or mol L–1s–1
1
R® P
d[R]/dt = -k[R]
[R] = [R]0e-kt
ln[R] vs t
ln 2/k
time-1 or s–1
or kt =
ln{[R]0/[R]}
Rationalised 2023-24
78
Chemistry
Most of the chemical reactions are accelerated by increase in temperature For example, in decomposition of N2O5, the time taken for half of the
original amount of material to decompose is 12 min at 50oC, 5 h at
25oC and 10 days at 0oC You also know that in a mixture of potassium
permanganate (KMnO4) and oxalic acid (H2C2O4), potassium
permanganate gets decolourised faster at a higher temperature than
that at a lower temperature It has been found that for a chemical reaction with rise in
temperature by 10°, the rate constant is nearly doubled |
1 | 3272-3275 | For example, in decomposition of N2O5, the time taken for half of the
original amount of material to decompose is 12 min at 50oC, 5 h at
25oC and 10 days at 0oC You also know that in a mixture of potassium
permanganate (KMnO4) and oxalic acid (H2C2O4), potassium
permanganate gets decolourised faster at a higher temperature than
that at a lower temperature It has been found that for a chemical reaction with rise in
temperature by 10°, the rate constant is nearly doubled The temperature dependence of the rate of a chemical reaction can
be accurately explained by Arrhenius equation (3 |
1 | 3273-3276 | You also know that in a mixture of potassium
permanganate (KMnO4) and oxalic acid (H2C2O4), potassium
permanganate gets decolourised faster at a higher temperature than
that at a lower temperature It has been found that for a chemical reaction with rise in
temperature by 10°, the rate constant is nearly doubled The temperature dependence of the rate of a chemical reaction can
be accurately explained by Arrhenius equation (3 18) |
1 | 3274-3277 | It has been found that for a chemical reaction with rise in
temperature by 10°, the rate constant is nearly doubled The temperature dependence of the rate of a chemical reaction can
be accurately explained by Arrhenius equation (3 18) It was first
proposed by Dutch chemist, J |
1 | 3275-3278 | The temperature dependence of the rate of a chemical reaction can
be accurately explained by Arrhenius equation (3 18) It was first
proposed by Dutch chemist, J H |
1 | 3276-3279 | 18) It was first
proposed by Dutch chemist, J H van’t Hoff but Swedish chemist,
Arrhenius provided its physical justification and interpretation |
1 | 3277-3280 | It was first
proposed by Dutch chemist, J H van’t Hoff but Swedish chemist,
Arrhenius provided its physical justification and interpretation 3 |
1 | 3278-3281 | H van’t Hoff but Swedish chemist,
Arrhenius provided its physical justification and interpretation 3 4 Temperature
3 |
1 | 3279-3282 | van’t Hoff but Swedish chemist,
Arrhenius provided its physical justification and interpretation 3 4 Temperature
3 4 Temperature
3 |
1 | 3280-3283 | 3 4 Temperature
3 4 Temperature
3 4 Temperature
3 |
1 | 3281-3284 | 4 Temperature
3 4 Temperature
3 4 Temperature
3 4 Temperature
3 |
1 | 3282-3285 | 4 Temperature
3 4 Temperature
3 4 Temperature
3 4 Temperature
Dependence of
Dependence of
Dependence of
Dependence of
Dependence of
the Rate of a
the Rate of a
the Rate of a
the Rate of a
the Rate of a
Reaction
Reaction
Reaction
Reaction
Reaction
The order of a reaction is sometimes altered by conditions |
1 | 3283-3286 | 4 Temperature
3 4 Temperature
3 4 Temperature
Dependence of
Dependence of
Dependence of
Dependence of
Dependence of
the Rate of a
the Rate of a
the Rate of a
the Rate of a
the Rate of a
Reaction
Reaction
Reaction
Reaction
Reaction
The order of a reaction is sometimes altered by conditions There
are many reactions which obey first order rate law although they are
higher order reactions |
1 | 3284-3287 | 4 Temperature
3 4 Temperature
Dependence of
Dependence of
Dependence of
Dependence of
Dependence of
the Rate of a
the Rate of a
the Rate of a
the Rate of a
the Rate of a
Reaction
Reaction
Reaction
Reaction
Reaction
The order of a reaction is sometimes altered by conditions There
are many reactions which obey first order rate law although they are
higher order reactions Consider the hydrolysis of ethyl acetate which
is a chemical reaction between ethyl acetate and water |
1 | 3285-3288 | 4 Temperature
Dependence of
Dependence of
Dependence of
Dependence of
Dependence of
the Rate of a
the Rate of a
the Rate of a
the Rate of a
the Rate of a
Reaction
Reaction
Reaction
Reaction
Reaction
The order of a reaction is sometimes altered by conditions There
are many reactions which obey first order rate law although they are
higher order reactions Consider the hydrolysis of ethyl acetate which
is a chemical reaction between ethyl acetate and water In reality, it
is a second order reaction and concentration of both ethyl acetate and
water affect the rate of the reaction |
1 | 3286-3289 | There
are many reactions which obey first order rate law although they are
higher order reactions Consider the hydrolysis of ethyl acetate which
is a chemical reaction between ethyl acetate and water In reality, it
is a second order reaction and concentration of both ethyl acetate and
water affect the rate of the reaction But water is taken in large excess
for hydrolysis, therefore, concentration of water is not altered much
during the reaction |
1 | 3287-3290 | Consider the hydrolysis of ethyl acetate which
is a chemical reaction between ethyl acetate and water In reality, it
is a second order reaction and concentration of both ethyl acetate and
water affect the rate of the reaction But water is taken in large excess
for hydrolysis, therefore, concentration of water is not altered much
during the reaction Thus, the rate of reaction is affected by
concentration of ethyl acetate only |
1 | 3288-3291 | In reality, it
is a second order reaction and concentration of both ethyl acetate and
water affect the rate of the reaction But water is taken in large excess
for hydrolysis, therefore, concentration of water is not altered much
during the reaction Thus, the rate of reaction is affected by
concentration of ethyl acetate only For example, during the hydrolysis
of 0 |
1 | 3289-3292 | But water is taken in large excess
for hydrolysis, therefore, concentration of water is not altered much
during the reaction Thus, the rate of reaction is affected by
concentration of ethyl acetate only For example, during the hydrolysis
of 0 01 mol of ethyl acetate with 10 mol of water, amounts of the
reactants and products at the beginning (t = 0) and completion (t) of
the reaction are given as under |
1 | 3290-3293 | Thus, the rate of reaction is affected by
concentration of ethyl acetate only For example, during the hydrolysis
of 0 01 mol of ethyl acetate with 10 mol of water, amounts of the
reactants and products at the beginning (t = 0) and completion (t) of
the reaction are given as under CH3COOC2H5 + H2O
H
CH3COOH + C2H5OH
t = 0
0 |
1 | 3291-3294 | For example, during the hydrolysis
of 0 01 mol of ethyl acetate with 10 mol of water, amounts of the
reactants and products at the beginning (t = 0) and completion (t) of
the reaction are given as under CH3COOC2H5 + H2O
H
CH3COOH + C2H5OH
t = 0
0 01 mol
10 mol
0 mol
0 mol
t
0 mol
9 |
1 | 3292-3295 | 01 mol of ethyl acetate with 10 mol of water, amounts of the
reactants and products at the beginning (t = 0) and completion (t) of
the reaction are given as under CH3COOC2H5 + H2O
H
CH3COOH + C2H5OH
t = 0
0 01 mol
10 mol
0 mol
0 mol
t
0 mol
9 99 mol
0 |
1 | 3293-3296 | CH3COOC2H5 + H2O
H
CH3COOH + C2H5OH
t = 0
0 01 mol
10 mol
0 mol
0 mol
t
0 mol
9 99 mol
0 01 mol
0 |
1 | 3294-3297 | 01 mol
10 mol
0 mol
0 mol
t
0 mol
9 99 mol
0 01 mol
0 01 mol
The concentration of water does not get altered much during the
course of the reaction |
1 | 3295-3298 | 99 mol
0 01 mol
0 01 mol
The concentration of water does not get altered much during the
course of the reaction So, the reaction behaves as first order reaction |
1 | 3296-3299 | 01 mol
0 01 mol
The concentration of water does not get altered much during the
course of the reaction So, the reaction behaves as first order reaction Such reactions are called pseudo first order reactions |
1 | 3297-3300 | 01 mol
The concentration of water does not get altered much during the
course of the reaction So, the reaction behaves as first order reaction Such reactions are called pseudo first order reactions Inversion of cane sugar is another pseudo first order reaction |
1 | 3298-3301 | So, the reaction behaves as first order reaction Such reactions are called pseudo first order reactions Inversion of cane sugar is another pseudo first order reaction C12H22O11 + H2O
H+
→ C6H12O6 + C6H12O6
Cane sugar
Glucose Fructose
Rate = k [C12H22O11]
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
3 |
1 | 3299-3302 | Such reactions are called pseudo first order reactions Inversion of cane sugar is another pseudo first order reaction C12H22O11 + H2O
H+
→ C6H12O6 + C6H12O6
Cane sugar
Glucose Fructose
Rate = k [C12H22O11]
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
3 5
A first order reaction has a rate constant 1 |
1 | 3300-3303 | Inversion of cane sugar is another pseudo first order reaction C12H22O11 + H2O
H+
→ C6H12O6 + C6H12O6
Cane sugar
Glucose Fructose
Rate = k [C12H22O11]
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
3 5
A first order reaction has a rate constant 1 15 × 10-3 s-1 |
1 | 3301-3304 | C12H22O11 + H2O
H+
→ C6H12O6 + C6H12O6
Cane sugar
Glucose Fructose
Rate = k [C12H22O11]
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
3 5
A first order reaction has a rate constant 1 15 × 10-3 s-1 How long will 5 g
of this reactant take to reduce to 3 g |
1 | 3302-3305 | 5
A first order reaction has a rate constant 1 15 × 10-3 s-1 How long will 5 g
of this reactant take to reduce to 3 g 3 |
1 | 3303-3306 | 15 × 10-3 s-1 How long will 5 g
of this reactant take to reduce to 3 g 3 6
Time required to decompose SO2Cl2 to half of its initial amount is 60
minutes |
1 | 3304-3307 | How long will 5 g
of this reactant take to reduce to 3 g 3 6
Time required to decompose SO2Cl2 to half of its initial amount is 60
minutes If the decomposition is a first order reaction, calculate the rate
constant of the reaction |
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