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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)
New answer posted
a year 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
a year 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
a year agoNew answer posted
a year 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
New question posted
a year agoNew answer posted
a year agoContributor-Level 10
7.70
As Bond dissociation energy generally decreases on moving down the group as the atomic size of the element increases. However, among halogens, the bond dissociation energy of F2 is lower than that of Cl2 and F2 due to the small atomic size of
Thus increasing order for bond dissociation energy among halogens is as follows:I22
As Bond dissociation energy of H-X molecules where X is the halogen decreases with increase in the atomic HI is the strongest acid as it loses H atom easily due to weak bonding between H and I.
So Increasing acid strength is as follows: HF
Basic strength decreases as we move from Nitrogen to Bismuth down the group
New answer posted
a year agoContributor-Level 10
7.69
(i) XeO3 can be produced by hydrolysis of XeF4 and XeF6 under controlled pH of the medium in which reaction is taking place as shown below:
6XeF4 + 12H2O → 4Xe + 2XeO3 + 24HF + 3O2
XeF6 + 3H2O → XeO3 + 6HF
(ii) XeOF4 can be obtained on partial hydrolysis of XeF6 as shown below:
XeF6 + H2O → XeOF4 + 2HF
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