Chemical Kinetics
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All energetically effective collisions do not result in a chemical change. Explain with the help of an example.
All energetically effective collisions do not result in a chemical change. Explain with the help of an example.
This is a Long Answer Type Question as classified in NCERT Exemplar
The reaction rate in collision theory is determined by two factors: energy and orientation factor. Certain reactions can be highly exothermic and energetically favoured, meaning the reactants have enough activation energy to collide
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4.39 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
4.39 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
4.39 Given, k2 = 4k1, T1 = 293K and T2 = 313K
We know that from the Arrhenius equation, we obtain
On solving, we get,
Ea = 58263.33 J mol-1 or 58.26 kJ mol-1
Upvote 4.38 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s –1, calculate k at 318K and Ea.
4.38 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s –1, calculate k at 318K and Ea.
4.38 We know, time t = (2.303/k) × log ( [R]0/ [R])
Where, k- rate constant
[R]0-Initial concentration
[R]-Concentration at time ‘t’
At 298K, If 10% is completed, then 90% is remaining. t = (2.303/k) × log ( [R]0/0.9 [R]0)
t = (2.303/k) × log (1/0.9) t = 0.1054 / k
At temperature 308K, 25% is comp
Upvote 4.37 The decomposition of A into product has value of k as 4.5× 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1?
4.37 The decomposition of A into product has value of k as 4.5× 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1?
4.37 From Arrhenius equation, we obtain
Also, k1 = 4.5 × 103 s -1
T1 = 273 + 10 = 283 K
k2 = 1.5 × 104 s -1
Ea = 60 kJ mol -1 = 6.0 × 104 J mol -1
Then,
→ 0.5229 = 3133.627 × (T2-283)/ (283 × T2)
→ 0.0472T2 = T2-283 T2 = 297K or T2 = 240 C
Upvote 4.36 The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
4.36 The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
4.36 We know, The Arrhenius equation is given by k = Ae-Ea/RT Taking natural log on both sides,
Ln k = ln A- (Ea/RT)
Thus, log k = log A - (Ea/2.303RT). eqn 1
The given equation is log k = 14.34 – 1.25 * 104K/T. eqn 2
Comparing 2 equations, Ea/2.303R = 1.25 * 104K
Ea = 1.25 * 104K * 2.303 * 8.314
Ea
Upvote 4.35 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/T. Calculate Ea
4.35 The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/T. Calculate Ea
4.35 The given equation is
k = (4.5 x 1011 s-1) e-28000K/T (i)
Comparing, Arrhenius equation
k = Ae -E/RT (ii)
We get, Ea / RT = 28000K / T
⇒Ea = R x 28000K
= 8.314 J K-1mol-1 × 28000 K
= 232792 J mol–1 or 232.792 kJ mol–1
Upvote 4.34 Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t 1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
4.34 Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t 1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
4.34 t1/2 = 3.00 hours
We know, t1/2 = 0.693/k
? k = 0.693/3 k = 0.231 hrs-1
We know, time
Where, k- rate constant
[R]° -Initial concentration
[R]-Concentration at time 't'
Thus, substituting the values,
log ( [R]0/ [R]) = 0.8
log ( [R]/ [R]0) = -0.8
[R]/ [R]0 = 0.158
Hence, 0.158 fraction of
Upvote 4.33 Consider a certain reaction A → Products with k = 2.0 × 10 –2s –1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L–1
4.33 Consider a certain reaction A → Products with k = 2.0 × 10 –2s –1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L–1
4.33 Given,
k = 2.0 * 10–2s-1
time t = 100s
Concentration [A0] = 1.0 mol L-1
We know,
On substituting the values, Log (1/ [A]) = 2.303/2
Log [A] = -2.303/2 [A] = 0.135 mol L–1
Upvote 4.32 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s – 1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
4.32 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s – 1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
4.32 Given,
k = 2.418 * 10-5 s-1
T = 546 K
Ea = 179.9 kJ mol-1 = 179.9 * 103J mol-1
The Arrhenius equation is given by k = Ae-Ea/RT Taking natural log on both sides,
Ln k = ln A- (Ea/RT) Substituting the values,
ln (2.418 * 10-5 ) = ln A-179.9/ (8.314 * 546)
ln A = 12.5917
A = 3.9 * 1012 s-1 (approx
Upvote 4.31 The rate constant for the decomposition of N2O5 at various temperatures is given below:
Draw a graph between ln k and 1/T and calculate the values of A and Ea . Predict the rate constant at 30° and 50°C.
4.31 The rate constant for the decomposition of N2O5 at various temperatures is given below:
Draw a graph between ln k and 1/T and calculate the values of A and Ea . Predict the rate constant at 30° and 50°C.
4.31 To convert the temperature in °C to °K we add 273 K.
The graph is given as:
The Arrhenius equation is given by k = Ae-Ea/RT
Where, k- Rate constant
A- Constant
Ea-Activation Energy
R- Gas constant
T-Temperature
Taking natural log on both sides,
ln k = ln A- (Ea/RT). equation 1
By plotting a graph,
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