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New answer posted
a year agoContributor-Level 10
(d) Heavier elementsdo not form pπ– pπ bonds because their atomicorbitals are too large and diffuse to haveeffective overlapping.
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a year agoNew answer posted
a year 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
a year 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
a year 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
a year 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
a year 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
a year 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
a year 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
a year 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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