Chemical Kinetics
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New answer posted
4 months agoContributor-Level 10
This is a Long Answer Type Question as classified in NCERT Exemplar
Ans:
As illustrated in the graph, as the temperature rises, the peak pushes ahead, increasing probable kinetic energy while decreasing the number of molecules utilizing it, resulting in a faster rate of reaction.
New answer posted
4 months agoContributor-Level 10
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 effectively. However, they do not proceed at a fixed temperature, which is why the reactant molecules are not orientated properly during collisions, resulting in atoms of reactant molecules combining to produce products that do not face each other.
New answer posted
4 months agoContributor-Level 10
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
New answer posted
4 months agoContributor-Level 10
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 completed, 75% is remaining t' = (2.303/k') * log ( [R]0/0.75 [R]0)
t' = (2.303/k') * log (1/0.75) t' = 2.2877 / k'
But, t = t'
0.1054 / k = 2.2877 / k' / k = 2.7296
From Arrhenius equation, we obtain log k2/k1 = (Ea / 2.303 R) * (T2 - T1) / T1T2
Substituting the values,
Ea = 76640.09 J mol-1 or 76.64 kJ mol-1 We know, log k = log A –Ea/RT
Log k =
New answer posted
4 months agoContributor-Level 10
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
New answer posted
4 months agoContributor-Level 10
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 = 239339.3 J mol-1 (approximately) Ea = 239.34 kJ mol-1
Also, when t1/2 = 256 minutes,
k = 0.693 / t1/2
= 0.693 / 256
= 2.707 * 10-3 min-1 k = 4.51 * 10-5s–1
Substitute k = 4.51 * 10-5s–1 in eqn 2,
log 4.51 * 10-5 s–1 = 14.34 – 1.25 * 104K/T
log (0.654-5) = 14.34– 1.25 * 104K/T = 1.25 * 104/ [ 14.34- log (0.654-5)] T = 668.9K or T =
New answer posted
4 months agoContributor-Level 10
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
New answer posted
4 months agoContributor-Level 10
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 sucrose remains.
New answer posted
4 months agoContributor-Level 10
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
New answer posted
4 months agoContributor-Level 10
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 (approximately)
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