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New answer posted
8 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
8 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
8 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
8 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
8 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
8 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)
New answer posted
8 months ago
Contributor-Level 10
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, ln K Vs 1/T, we get y-intercept as ln A and Slope is –Ea/R.
Slope = (y2-y1)/ (x2-x1)
By substituting the values, slope = -12.301
? –Ea/R = -12.301
But, R = 8.314 JK-1mol-1
? aE= 8.314 JK-1mol-1 * 12.301 K
? aE= 102.27 kJ mol-1
Substituting the values in equation 1 for data at T = 273K
(? At T = 273K, ln k =-7.147)
On solving, we get ln A = 37.
New answer posted
8 months ago
Contributor-Level 10
4.30 When t = 0, the total partial pressure is P0 = 0.5 atm
When time t = t, the total partial pressure is Pt = P0 + p
P0-p = Pt-2p, but by the above equation, we know p = Pt-P0
Hence, P0-p = Pt-2 (Pt-P0)
Thus, P0-p = 2P0 – Pt
We know that time
t= 2.303/K log R0 / R
Where, k- rate constant
[R]° -Initial concentration of reactant [R]-Concentration of reactant at time 't'
Here concentration can be replaced by the corresponding partial pressures.
Hence, the equation becomes,
t= 2.303/K log P0 / P0 - P
t= 2.303/K log P0 / 2P0 - Pt
? equation 1
At time t = 100 s, Pt = 0.6 atm and P0 = 0.5 atm,
Substituting in equation 1,
100 = 2.303/k log 0.5 /
New question posted
8 months agoNew answer posted
8 months ago
Contributor-Level 10
4.29 When t = 0, the total partial pressure is P0 = 35.0 mm of Hg
When time t = t, the total partial pressure is Pt = P0 + p
P0-p = Pt-2p, but by the above equation, we know p = Pt-P0 Hence, P0-p = Pt-2 ( Pt-P0)
Thus, P0-p = 2P0 – Pt
We know that time
t= 2.303/K log R0 / R
Where, k- rate constant
[R]° -Initial concentration of reactant
[R]-Concentration of reactant at time 't'
Here concentration can be replaced by the corresponding partial pressures.
Hence, the equation becomes,
t= 2.303/K log P0 / P0 - P
t= 2.303/K log P0 / 2P0 - Pt
? equation 1
At time t = 360 s, Pt = 54 mm of Hg and P0 = 30 mm of Hg, Substituting in equation 1,
360 = 2
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