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1 | 3105-3108 | 3 3: Variation in the concentration
vs time plot for a zero order
reaction
Time
k = -slope
Concentration of R
[R ]
0
0
Comparing (3 6) with equation of a straight line,
y = mx + c, if we plot [R] against t, we get a straight
line (Fig 3 |
1 | 3106-3109 | 3: Variation in the concentration
vs time plot for a zero order
reaction
Time
k = -slope
Concentration of R
[R ]
0
0
Comparing (3 6) with equation of a straight line,
y = mx + c, if we plot [R] against t, we get a straight
line (Fig 3 3) with slope = –k and intercept equal
to [R]0 |
1 | 3107-3110 | 6) with equation of a straight line,
y = mx + c, if we plot [R] against t, we get a straight
line (Fig 3 3) with slope = –k and intercept equal
to [R]0 Further simplifying equation (3 |
1 | 3108-3111 | 3 3) with slope = –k and intercept equal
to [R]0 Further simplifying equation (3 6), we get the rate
constant, k as
[
]
[ ]
R0
R
k
t
−
=
(3 |
1 | 3109-3112 | 3) with slope = –k and intercept equal
to [R]0 Further simplifying equation (3 6), we get the rate
constant, k as
[
]
[ ]
R0
R
k
t
−
=
(3 7)
Zero order reactions are relatively uncommon but
they occur under special conditions |
1 | 3110-3113 | Further simplifying equation (3 6), we get the rate
constant, k as
[
]
[ ]
R0
R
k
t
−
=
(3 7)
Zero order reactions are relatively uncommon but
they occur under special conditions Some enzyme
catalysed reactions and reactions which occur on
metal surfaces are a few examples of zero order
reactions |
1 | 3111-3114 | 6), we get the rate
constant, k as
[
]
[ ]
R0
R
k
t
−
=
(3 7)
Zero order reactions are relatively uncommon but
they occur under special conditions Some enzyme
catalysed reactions and reactions which occur on
metal surfaces are a few examples of zero order
reactions The decomposition of gaseous ammonia
on a hot platinum surface is a zero order reaction at
high pressure |
1 | 3112-3115 | 7)
Zero order reactions are relatively uncommon but
they occur under special conditions Some enzyme
catalysed reactions and reactions which occur on
metal surfaces are a few examples of zero order
reactions The decomposition of gaseous ammonia
on a hot platinum surface is a zero order reaction at
high pressure ( )
( )
( )
1130K
3
2
2
Pt catalyst
2NH
g
N
g +3H
g
→
Rate = k [NH3]0 = k
In this reaction, platinum metal acts as a catalyst |
1 | 3113-3116 | Some enzyme
catalysed reactions and reactions which occur on
metal surfaces are a few examples of zero order
reactions The decomposition of gaseous ammonia
on a hot platinum surface is a zero order reaction at
high pressure ( )
( )
( )
1130K
3
2
2
Pt catalyst
2NH
g
N
g +3H
g
→
Rate = k [NH3]0 = k
In this reaction, platinum metal acts as a catalyst At high pressure,
the metal surface gets saturated with gas molecules |
1 | 3114-3117 | The decomposition of gaseous ammonia
on a hot platinum surface is a zero order reaction at
high pressure ( )
( )
( )
1130K
3
2
2
Pt catalyst
2NH
g
N
g +3H
g
→
Rate = k [NH3]0 = k
In this reaction, platinum metal acts as a catalyst At high pressure,
the metal surface gets saturated with gas molecules So, a further
change in reaction conditions is unable to alter the amount of ammonia
on the surface of the catalyst making rate of the reaction independent
of its concentration |
1 | 3115-3118 | ( )
( )
( )
1130K
3
2
2
Pt catalyst
2NH
g
N
g +3H
g
→
Rate = k [NH3]0 = k
In this reaction, platinum metal acts as a catalyst At high pressure,
the metal surface gets saturated with gas molecules So, a further
change in reaction conditions is unable to alter the amount of ammonia
on the surface of the catalyst making rate of the reaction independent
of its concentration The thermal decomposition of HI on gold surface
is another example of zero order reaction |
1 | 3116-3119 | At high pressure,
the metal surface gets saturated with gas molecules So, a further
change in reaction conditions is unable to alter the amount of ammonia
on the surface of the catalyst making rate of the reaction independent
of its concentration The thermal decomposition of HI on gold surface
is another example of zero order reaction In this class of reactions, the rate of the reaction is proportional to the
first power of the concentration of the reactant R |
1 | 3117-3120 | So, a further
change in reaction conditions is unable to alter the amount of ammonia
on the surface of the catalyst making rate of the reaction independent
of its concentration The thermal decomposition of HI on gold surface
is another example of zero order reaction In this class of reactions, the rate of the reaction is proportional to the
first power of the concentration of the reactant R For example,
R ® P
[
]
[
]
d R
Rate
R
d
k
t
= −
=
or
[ ]
[ ]
d R
– d
R
k t
=
Integrating this equation, we get
ln [R] = – kt + I
(3 |
1 | 3118-3121 | The thermal decomposition of HI on gold surface
is another example of zero order reaction In this class of reactions, the rate of the reaction is proportional to the
first power of the concentration of the reactant R For example,
R ® P
[
]
[
]
d R
Rate
R
d
k
t
= −
=
or
[ ]
[ ]
d R
– d
R
k t
=
Integrating this equation, we get
ln [R] = – kt + I
(3 8)
Again, I is the constant of integration and its value can be determined
easily |
1 | 3119-3122 | In this class of reactions, the rate of the reaction is proportional to the
first power of the concentration of the reactant R For example,
R ® P
[
]
[
]
d R
Rate
R
d
k
t
= −
=
or
[ ]
[ ]
d R
– d
R
k t
=
Integrating this equation, we get
ln [R] = – kt + I
(3 8)
Again, I is the constant of integration and its value can be determined
easily When t = 0, R = [R]0, where [R]0 is the initial concentration of the
reactant |
1 | 3120-3123 | For example,
R ® P
[
]
[
]
d R
Rate
R
d
k
t
= −
=
or
[ ]
[ ]
d R
– d
R
k t
=
Integrating this equation, we get
ln [R] = – kt + I
(3 8)
Again, I is the constant of integration and its value can be determined
easily When t = 0, R = [R]0, where [R]0 is the initial concentration of the
reactant Therefore, equation (3 |
1 | 3121-3124 | 8)
Again, I is the constant of integration and its value can be determined
easily When t = 0, R = [R]0, where [R]0 is the initial concentration of the
reactant Therefore, equation (3 8) can be written as
ln [R]0 = –k × 0 + I
ln [R]0 = I
Substituting the value of I in equation (3 |
1 | 3122-3125 | When t = 0, R = [R]0, where [R]0 is the initial concentration of the
reactant Therefore, equation (3 8) can be written as
ln [R]0 = –k × 0 + I
ln [R]0 = I
Substituting the value of I in equation (3 8)
ln[R] = –kt + ln[R]0
(3 |
1 | 3123-3126 | Therefore, equation (3 8) can be written as
ln [R]0 = –k × 0 + I
ln [R]0 = I
Substituting the value of I in equation (3 8)
ln[R] = –kt + ln[R]0
(3 9)
3 |
1 | 3124-3127 | 8) can be written as
ln [R]0 = –k × 0 + I
ln [R]0 = I
Substituting the value of I in equation (3 8)
ln[R] = –kt + ln[R]0
(3 9)
3 3 |
1 | 3125-3128 | 8)
ln[R] = –kt + ln[R]0
(3 9)
3 3 2 First Order
Reactions
Rationalised 2023-24
73
Chemical Kinetics
Rearranging this equation
[
]
[
]0
lnR
R
kt
= −
or
1 lnR0
R
k
t
(3 |
1 | 3126-3129 | 9)
3 3 2 First Order
Reactions
Rationalised 2023-24
73
Chemical Kinetics
Rearranging this equation
[
]
[
]0
lnR
R
kt
= −
or
1 lnR0
R
k
t
(3 10)
At time t1 from equation (3 |
1 | 3127-3130 | 3 2 First Order
Reactions
Rationalised 2023-24
73
Chemical Kinetics
Rearranging this equation
[
]
[
]0
lnR
R
kt
= −
or
1 lnR0
R
k
t
(3 10)
At time t1 from equation (3 8)
*ln[R]1 = – kt1 + *ln[R]0
(3 |
1 | 3128-3131 | 2 First Order
Reactions
Rationalised 2023-24
73
Chemical Kinetics
Rearranging this equation
[
]
[
]0
lnR
R
kt
= −
or
1 lnR0
R
k
t
(3 10)
At time t1 from equation (3 8)
*ln[R]1 = – kt1 + *ln[R]0
(3 11)
At time t2
ln[R]2 = – kt2 + ln[R]0
(3 |
1 | 3129-3132 | 10)
At time t1 from equation (3 8)
*ln[R]1 = – kt1 + *ln[R]0
(3 11)
At time t2
ln[R]2 = – kt2 + ln[R]0
(3 12)
where, [R]1 and [R]2 are the concentrations of the reactants at time
t1 and t2 respectively |
1 | 3130-3133 | 8)
*ln[R]1 = – kt1 + *ln[R]0
(3 11)
At time t2
ln[R]2 = – kt2 + ln[R]0
(3 12)
where, [R]1 and [R]2 are the concentrations of the reactants at time
t1 and t2 respectively Subtracting (3 |
1 | 3131-3134 | 11)
At time t2
ln[R]2 = – kt2 + ln[R]0
(3 12)
where, [R]1 and [R]2 are the concentrations of the reactants at time
t1 and t2 respectively Subtracting (3 12) from (3 |
1 | 3132-3135 | 12)
where, [R]1 and [R]2 are the concentrations of the reactants at time
t1 and t2 respectively Subtracting (3 12) from (3 11)
ln[R]1– ln[R]2 = – kt1 – (–kt2)
[
]
[ ]
(
)
−
=
1
2
1
2
lnR
R
t
t
k
(
)
[
]
[
]
1
2
1
2
1
R
ln
R
k
t
t
=
−
(3 |
1 | 3133-3136 | Subtracting (3 12) from (3 11)
ln[R]1– ln[R]2 = – kt1 – (–kt2)
[
]
[ ]
(
)
−
=
1
2
1
2
lnR
R
t
t
k
(
)
[
]
[
]
1
2
1
2
1
R
ln
R
k
t
t
=
−
(3 13)
Equation (3 |
1 | 3134-3137 | 12) from (3 11)
ln[R]1– ln[R]2 = – kt1 – (–kt2)
[
]
[ ]
(
)
−
=
1
2
1
2
lnR
R
t
t
k
(
)
[
]
[
]
1
2
1
2
1
R
ln
R
k
t
t
=
−
(3 13)
Equation (3 9) can also be written as
[[ ]
]
= −
0
R
ln
R
kt
Taking antilog of both sides
[R] = [R]0 e–kt
(3 |
1 | 3135-3138 | 11)
ln[R]1– ln[R]2 = – kt1 – (–kt2)
[
]
[ ]
(
)
−
=
1
2
1
2
lnR
R
t
t
k
(
)
[
]
[
]
1
2
1
2
1
R
ln
R
k
t
t
=
−
(3 13)
Equation (3 9) can also be written as
[[ ]
]
= −
0
R
ln
R
kt
Taking antilog of both sides
[R] = [R]0 e–kt
(3 14)
Comparing equation (3 |
1 | 3136-3139 | 13)
Equation (3 9) can also be written as
[[ ]
]
= −
0
R
ln
R
kt
Taking antilog of both sides
[R] = [R]0 e–kt
(3 14)
Comparing equation (3 9) with y = mx + c, if we plot ln [R] against
t (Fig |
1 | 3137-3140 | 9) can also be written as
[[ ]
]
= −
0
R
ln
R
kt
Taking antilog of both sides
[R] = [R]0 e–kt
(3 14)
Comparing equation (3 9) with y = mx + c, if we plot ln [R] against
t (Fig 3 |
1 | 3138-3141 | 14)
Comparing equation (3 9) with y = mx + c, if we plot ln [R] against
t (Fig 3 4) we get a straight line with slope = –k and intercept equal to
ln [R]0
The first order rate equation (3 |
1 | 3139-3142 | 9) with y = mx + c, if we plot ln [R] against
t (Fig 3 4) we get a straight line with slope = –k and intercept equal to
ln [R]0
The first order rate equation (3 10) can also be written in the form
[ ]
[ ]
2 |
1 | 3140-3143 | 3 4) we get a straight line with slope = –k and intercept equal to
ln [R]0
The first order rate equation (3 10) can also be written in the form
[ ]
[ ]
2 303 logR0
R
k
t
=
(3 |
1 | 3141-3144 | 4) we get a straight line with slope = –k and intercept equal to
ln [R]0
The first order rate equation (3 10) can also be written in the form
[ ]
[ ]
2 303 logR0
R
k
t
=
(3 15)
*
[
]
[
]
logR0
2 |
1 | 3142-3145 | 10) can also be written in the form
[ ]
[ ]
2 303 logR0
R
k
t
=
(3 15)
*
[
]
[
]
logR0
2 303
R
kt
=
If we plot a graph between log [R]0/[R] vs t, (Fig |
1 | 3143-3146 | 303 logR0
R
k
t
=
(3 15)
*
[
]
[
]
logR0
2 303
R
kt
=
If we plot a graph between log [R]0/[R] vs t, (Fig 3 |
1 | 3144-3147 | 15)
*
[
]
[
]
logR0
2 303
R
kt
=
If we plot a graph between log [R]0/[R] vs t, (Fig 3 5),
the slope = k/2 |
1 | 3145-3148 | 303
R
kt
=
If we plot a graph between log [R]0/[R] vs t, (Fig 3 5),
the slope = k/2 303
Hydrogenation of ethene is an example of first order reaction |
1 | 3146-3149 | 3 5),
the slope = k/2 303
Hydrogenation of ethene is an example of first order reaction C2H4(g) + H2 (g) ® C2H6(g)
Rate = k [C2H4]
All natural and artificial radioactive decay of unstable nuclei take
place by first order kinetics |
1 | 3147-3150 | 5),
the slope = k/2 303
Hydrogenation of ethene is an example of first order reaction C2H4(g) + H2 (g) ® C2H6(g)
Rate = k [C2H4]
All natural and artificial radioactive decay of unstable nuclei take
place by first order kinetics * Refer to Appendix-IV for ln and log (logarithms) |
1 | 3148-3151 | 303
Hydrogenation of ethene is an example of first order reaction C2H4(g) + H2 (g) ® C2H6(g)
Rate = k [C2H4]
All natural and artificial radioactive decay of unstable nuclei take
place by first order kinetics * Refer to Appendix-IV for ln and log (logarithms) Rationalised 2023-24
74
Chemistry
Fig |
1 | 3149-3152 | C2H4(g) + H2 (g) ® C2H6(g)
Rate = k [C2H4]
All natural and artificial radioactive decay of unstable nuclei take
place by first order kinetics * Refer to Appendix-IV for ln and log (logarithms) Rationalised 2023-24
74
Chemistry
Fig 3 |
1 | 3150-3153 | * Refer to Appendix-IV for ln and log (logarithms) Rationalised 2023-24
74
Chemistry
Fig 3 4: A plot between ln[R] and t
for a first order reaction
Fig |
1 | 3151-3154 | Rationalised 2023-24
74
Chemistry
Fig 3 4: A plot between ln[R] and t
for a first order reaction
Fig 3 |
1 | 3152-3155 | 3 4: A plot between ln[R] and t
for a first order reaction
Fig 3 5: Plot of log [R]0/[R] vs time for a
first order reaction
Slope =
/2 |
1 | 3153-3156 | 4: A plot between ln[R] and t
for a first order reaction
Fig 3 5: Plot of log [R]0/[R] vs time for a
first order reaction
Slope =
/2 303
k
log ([R /[R])
0]
Time
0
226
4
222
88
2
86
Ra
He
Rn
Rate = k [Ra]
Decomposition of N2O5 and N2O are some more examples of first
order reactions |
1 | 3154-3157 | 3 5: Plot of log [R]0/[R] vs time for a
first order reaction
Slope =
/2 303
k
log ([R /[R])
0]
Time
0
226
4
222
88
2
86
Ra
He
Rn
Rate = k [Ra]
Decomposition of N2O5 and N2O are some more examples of first
order reactions The initial concentration of N2O5 in the following first order reaction
N2O5(g) ® 2 NO2(g) + 1/2O2 (g) was 1 |
1 | 3155-3158 | 5: Plot of log [R]0/[R] vs time for a
first order reaction
Slope =
/2 303
k
log ([R /[R])
0]
Time
0
226
4
222
88
2
86
Ra
He
Rn
Rate = k [Ra]
Decomposition of N2O5 and N2O are some more examples of first
order reactions The initial concentration of N2O5 in the following first order reaction
N2O5(g) ® 2 NO2(g) + 1/2O2 (g) was 1 24 × 10–2 mol L–1 at 318 K |
1 | 3156-3159 | 303
k
log ([R /[R])
0]
Time
0
226
4
222
88
2
86
Ra
He
Rn
Rate = k [Ra]
Decomposition of N2O5 and N2O are some more examples of first
order reactions The initial concentration of N2O5 in the following first order reaction
N2O5(g) ® 2 NO2(g) + 1/2O2 (g) was 1 24 × 10–2 mol L–1 at 318 K The
concentration of N2O5 after 60 minutes was 0 |
1 | 3157-3160 | The initial concentration of N2O5 in the following first order reaction
N2O5(g) ® 2 NO2(g) + 1/2O2 (g) was 1 24 × 10–2 mol L–1 at 318 K The
concentration of N2O5 after 60 minutes was 0 20 × 10–2 mol L–1 |
1 | 3158-3161 | 24 × 10–2 mol L–1 at 318 K The
concentration of N2O5 after 60 minutes was 0 20 × 10–2 mol L–1 Calculate
the rate constant of the reaction at 318 K |
1 | 3159-3162 | The
concentration of N2O5 after 60 minutes was 0 20 × 10–2 mol L–1 Calculate
the rate constant of the reaction at 318 K For a first order reaction
1
2
R
log
R
=
2
1
2 |
1 | 3160-3163 | 20 × 10–2 mol L–1 Calculate
the rate constant of the reaction at 318 K For a first order reaction
1
2
R
log
R
=
2
1
2 303
k t
t
k
=
1
2
1
2
t2 |
1 | 3161-3164 | Calculate
the rate constant of the reaction at 318 K For a first order reaction
1
2
R
log
R
=
2
1
2 303
k t
t
k
=
1
2
1
2
t2 303 log R
t
R
=
2
1
2
1
1 |
1 | 3162-3165 | For a first order reaction
1
2
R
log
R
=
2
1
2 303
k t
t
k
=
1
2
1
2
t2 303 log R
t
R
=
2
1
2
1
1 24 10
mol L
2 |
1 | 3163-3166 | 303
k t
t
k
=
1
2
1
2
t2 303 log R
t
R
=
2
1
2
1
1 24 10
mol L
2 303
60 min 0 minlog
0 |
1 | 3164-3167 | 303 log R
t
R
=
2
1
2
1
1 24 10
mol L
2 303
60 min 0 minlog
0 20 10
mol L
=
1
2 |
1 | 3165-3168 | 24 10
mol L
2 303
60 min 0 minlog
0 20 10
mol L
=
1
2 303 log 6 |
1 | 3166-3169 | 303
60 min 0 minlog
0 20 10
mol L
=
1
2 303 log 6 2 min
60
k
= 0 |
1 | 3167-3170 | 20 10
mol L
=
1
2 303 log 6 2 min
60
k
= 0 0304 min-1
Example 3 |
1 | 3168-3171 | 303 log 6 2 min
60
k
= 0 0304 min-1
Example 3 5
Example 3 |
1 | 3169-3172 | 2 min
60
k
= 0 0304 min-1
Example 3 5
Example 3 5
Example 3 |
1 | 3170-3173 | 0304 min-1
Example 3 5
Example 3 5
Example 3 5
Example 3 |
1 | 3171-3174 | 5
Example 3 5
Example 3 5
Example 3 5
Example 3 |
1 | 3172-3175 | 5
Example 3 5
Example 3 5
Example 3 5
Solution
Solution
Solution
Solution
Solution
Let us consider a typical first order gas phase reaction
A(g) ® B(g) + C(g)
Let pi be the initial pressure of A and pt the total pressure at
time ‘t’ |
1 | 3173-3176 | 5
Example 3 5
Example 3 5
Solution
Solution
Solution
Solution
Solution
Let us consider a typical first order gas phase reaction
A(g) ® B(g) + C(g)
Let pi be the initial pressure of A and pt the total pressure at
time ‘t’ Integrated rate equation for such a reaction can be derived as
Total pressure pt = pA + pB + pC (pressure units)
Rationalised 2023-24
75
Chemical Kinetics
pA, pB and pC are the partial pressures of A, B and C, respectively |
1 | 3174-3177 | 5
Example 3 5
Solution
Solution
Solution
Solution
Solution
Let us consider a typical first order gas phase reaction
A(g) ® B(g) + C(g)
Let pi be the initial pressure of A and pt the total pressure at
time ‘t’ Integrated rate equation for such a reaction can be derived as
Total pressure pt = pA + pB + pC (pressure units)
Rationalised 2023-24
75
Chemical Kinetics
pA, pB and pC are the partial pressures of A, B and C, respectively If x atm be the decrease in pressure of A at time t and one mole each
of B and C is being formed, the increase in pressure of B and C will also
be x atm each |
1 | 3175-3178 | 5
Solution
Solution
Solution
Solution
Solution
Let us consider a typical first order gas phase reaction
A(g) ® B(g) + C(g)
Let pi be the initial pressure of A and pt the total pressure at
time ‘t’ Integrated rate equation for such a reaction can be derived as
Total pressure pt = pA + pB + pC (pressure units)
Rationalised 2023-24
75
Chemical Kinetics
pA, pB and pC are the partial pressures of A, B and C, respectively If x atm be the decrease in pressure of A at time t and one mole each
of B and C is being formed, the increase in pressure of B and C will also
be x atm each A(g) ® B(g) + C(g)
At t = 0
pi atm
0 atm
0 atm
At time t
(pi–x) atm
x atm
x atm
where, pi is the initial pressure at time t = 0 |
1 | 3176-3179 | Integrated rate equation for such a reaction can be derived as
Total pressure pt = pA + pB + pC (pressure units)
Rationalised 2023-24
75
Chemical Kinetics
pA, pB and pC are the partial pressures of A, B and C, respectively If x atm be the decrease in pressure of A at time t and one mole each
of B and C is being formed, the increase in pressure of B and C will also
be x atm each A(g) ® B(g) + C(g)
At t = 0
pi atm
0 atm
0 atm
At time t
(pi–x) atm
x atm
x atm
where, pi is the initial pressure at time t = 0 pt = (pi – x) + x + x = pi + x
x = (pt - pi)
where, pA = pi – x = pi – (pt – pi)
= 2pi – pt
k =
i
A
2 |
1 | 3177-3180 | If x atm be the decrease in pressure of A at time t and one mole each
of B and C is being formed, the increase in pressure of B and C will also
be x atm each A(g) ® B(g) + C(g)
At t = 0
pi atm
0 atm
0 atm
At time t
(pi–x) atm
x atm
x atm
where, pi is the initial pressure at time t = 0 pt = (pi – x) + x + x = pi + x
x = (pt - pi)
where, pA = pi – x = pi – (pt – pi)
= 2pi – pt
k =
i
A
2 303
log p
p
t
(3 |
1 | 3178-3181 | A(g) ® B(g) + C(g)
At t = 0
pi atm
0 atm
0 atm
At time t
(pi–x) atm
x atm
x atm
where, pi is the initial pressure at time t = 0 pt = (pi – x) + x + x = pi + x
x = (pt - pi)
where, pA = pi – x = pi – (pt – pi)
= 2pi – pt
k =
i
A
2 303
log p
p
t
(3 16)
=
i
i
t
2 |
1 | 3179-3182 | pt = (pi – x) + x + x = pi + x
x = (pt - pi)
where, pA = pi – x = pi – (pt – pi)
= 2pi – pt
k =
i
A
2 303
log p
p
t
(3 16)
=
i
i
t
2 303 log 2
p
p
p
t
The following data were obtained during the first order thermal
decomposition of N2O5 (g) at constant volume:
( )
( )
( )
2
5
2
4
2
g
g
g
2N O
2N O
O
→
+
S |
1 | 3180-3183 | 303
log p
p
t
(3 16)
=
i
i
t
2 303 log 2
p
p
p
t
The following data were obtained during the first order thermal
decomposition of N2O5 (g) at constant volume:
( )
( )
( )
2
5
2
4
2
g
g
g
2N O
2N O
O
→
+
S No |
1 | 3181-3184 | 16)
=
i
i
t
2 303 log 2
p
p
p
t
The following data were obtained during the first order thermal
decomposition of N2O5 (g) at constant volume:
( )
( )
( )
2
5
2
4
2
g
g
g
2N O
2N O
O
→
+
S No Time/s
Total Pressure/(atm)
1 |
1 | 3182-3185 | 303 log 2
p
p
p
t
The following data were obtained during the first order thermal
decomposition of N2O5 (g) at constant volume:
( )
( )
( )
2
5
2
4
2
g
g
g
2N O
2N O
O
→
+
S No Time/s
Total Pressure/(atm)
1 0
0 |
1 | 3183-3186 | No Time/s
Total Pressure/(atm)
1 0
0 5
2 |
1 | 3184-3187 | Time/s
Total Pressure/(atm)
1 0
0 5
2 100
0 |
1 | 3185-3188 | 0
0 5
2 100
0 512
Calculate the rate constant |
1 | 3186-3189 | 5
2 100
0 512
Calculate the rate constant Let the pressure of N2O5(g) decrease by 2x atm |
1 | 3187-3190 | 100
0 512
Calculate the rate constant Let the pressure of N2O5(g) decrease by 2x atm As two moles of
N2O5 decompose to give two moles of N2O4(g) and one mole of O2 (g),
the pressure of N2O4 (g) increases by 2x atm and that of O2 (g)
increases by x atm |
1 | 3188-3191 | 512
Calculate the rate constant Let the pressure of N2O5(g) decrease by 2x atm As two moles of
N2O5 decompose to give two moles of N2O4(g) and one mole of O2 (g),
the pressure of N2O4 (g) increases by 2x atm and that of O2 (g)
increases by x atm ( )
( )
( )
2
5
2
4
2
g
g
g
2N O
2N O
O
→
+
Start t = 0 0 |
1 | 3189-3192 | Let the pressure of N2O5(g) decrease by 2x atm As two moles of
N2O5 decompose to give two moles of N2O4(g) and one mole of O2 (g),
the pressure of N2O4 (g) increases by 2x atm and that of O2 (g)
increases by x atm ( )
( )
( )
2
5
2
4
2
g
g
g
2N O
2N O
O
→
+
Start t = 0 0 5 atm
0 atm
0 atm
At time t (0 |
1 | 3190-3193 | As two moles of
N2O5 decompose to give two moles of N2O4(g) and one mole of O2 (g),
the pressure of N2O4 (g) increases by 2x atm and that of O2 (g)
increases by x atm ( )
( )
( )
2
5
2
4
2
g
g
g
2N O
2N O
O
→
+
Start t = 0 0 5 atm
0 atm
0 atm
At time t (0 5 – 2x) atm
2x atm
x atm
pt =
2
5
2
4
2
N O
N O
O
p
p
p
= (0 |
1 | 3191-3194 | ( )
( )
( )
2
5
2
4
2
g
g
g
2N O
2N O
O
→
+
Start t = 0 0 5 atm
0 atm
0 atm
At time t (0 5 – 2x) atm
2x atm
x atm
pt =
2
5
2
4
2
N O
N O
O
p
p
p
= (0 5 – 2x) + 2x + x = 0 |
1 | 3192-3195 | 5 atm
0 atm
0 atm
At time t (0 5 – 2x) atm
2x atm
x atm
pt =
2
5
2
4
2
N O
N O
O
p
p
p
= (0 5 – 2x) + 2x + x = 0 5 + x
x
0 |
1 | 3193-3196 | 5 – 2x) atm
2x atm
x atm
pt =
2
5
2
4
2
N O
N O
O
p
p
p
= (0 5 – 2x) + 2x + x = 0 5 + x
x
0 5
tp
=
−
2
pN O5
= 0 |
1 | 3194-3197 | 5 – 2x) + 2x + x = 0 5 + x
x
0 5
tp
=
−
2
pN O5
= 0 5 – 2x
= 0 |
1 | 3195-3198 | 5 + x
x
0 5
tp
=
−
2
pN O5
= 0 5 – 2x
= 0 5 – 2 (pt – 0 |
1 | 3196-3199 | 5
tp
=
−
2
pN O5
= 0 5 – 2x
= 0 5 – 2 (pt – 0 5) = 1 |
1 | 3197-3200 | 5 – 2x
= 0 5 – 2 (pt – 0 5) = 1 5 – 2pt
At t = 100 s; pt = 0 |
1 | 3198-3201 | 5 – 2 (pt – 0 5) = 1 5 – 2pt
At t = 100 s; pt = 0 512 atm
Example 3 |
1 | 3199-3202 | 5) = 1 5 – 2pt
At t = 100 s; pt = 0 512 atm
Example 3 6
Example 3 |
1 | 3200-3203 | 5 – 2pt
At t = 100 s; pt = 0 512 atm
Example 3 6
Example 3 6
Example 3 |
1 | 3201-3204 | 512 atm
Example 3 6
Example 3 6
Example 3 6
Example 3 |
1 | 3202-3205 | 6
Example 3 6
Example 3 6
Example 3 6
Example 3 |
1 | 3203-3206 | 6
Example 3 6
Example 3 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 |
1 | 3204-3207 | 6
Example 3 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 |
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