The plot of log k vs 1/T helps to calculate
1. Energy of activation
2. Rate constant of reaction
3. Order of reaction
4. Energy of activation as well as the frequency factor
(4) logK= logA - Ea/2.303RT
comparing with y = mx+c
log A = intercept
-Ea/2.303RT = slope
Ea = slope x (-2.303R)
So, Ea (energy of activation ) and A (Frequancy factor) can be calculated from above plot.
Two reactions of the same order have equal Pre-exponential factors but their activation energies differ by 24.9 kJ/mol. Calculate the ratio between the rate constants (K2/K1) these reactions at 27°C :
(1) 3 x 104
(2) 1/3 x 10-4
(3) 1/3 x 104
(4) 3 x 10-4
(1) log10K1 = log10A - Ea1/2.303RT ----(1)
log10K2 = log10A - Ea2/2.303RT ----(2)
log10(K2/K1) = (Ea1 - Ea2)/2.303RT
= log10(K2/K1) = (24.9x1000) / (2.303x8.314x300)
= K2/K1 = 2.99x104 = 3 x 104
A first order reaction has a specific reaction rate of 10-2s-1. How much time will it take for 20 g of the reactant to reduce to 5 g?
(a) 238.6 s (b) 138.6 s (c) 346.5 s (d) 693.0 s
(b) For a first order reaction,
Rate constant (k) = 2.303/t . log(a/a-x)
where, a = initial concentration
a-x = concentration after time 't'
t= time in 'sec'
Given, a= 20 g, a-x = 5g, k=10-2
... t =2.303/10-2 . log(20/5) = 138.6 s
Half-life for the first order reaction,
t1/2/2 = 0.693/k = 0.693/10-2 = 69.3s
Two half-lives are required for the reduction of 20 g of reactant into 5g
The decomposition of phosphine (PH3) on tungsten at low pressure is a first-order reaction. It is because the
(a) rate is proportional to the surface coverage
(b) rate is inversely proportional to the surface coverage
(c) rate is independent of the surface coverage
(d) rate of decomposition is very slow
(a) PH3 P +3/2. H2
This is an example of surface catalysed unimolecular decomposition.
For the above reaction, rate is given as
Rate = k/1+
where, = partial pressure of absorbing substrate.
At low pressure,
The rate of a first-order reaction is 0.04 mol L-1 s-1 at 10 sec and 0.03 mol L-1 s-1 at 20 sec after initiation of the reaction. The half-life period of the reaction is
(a) 34.1 s (b) 44.1 s (c) 54.1 s (d) 24.1 s
(d) Given, order of reaction = 1
Rate of reaction at 10 s = 0.04 mol L-1 s-1
Rate of reaction at 20 s = 0.03 mol L-1s-1
Half-lite period (t1/2) = ?
We have the equation for rate-constant 'k' in first
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R=8.314 J mol-1 K-1)
(a) 342 kJ mol-1
(b) 269 kJ mol-1
(c) 34.7 kJ mol-1
(d) 15.1 kJ mol-1
(c) Given, initial temperature,
R=8.314 J mol-1 K-1
Since, rate becomes double on raising temperature,
∴ r2=2r1 or r2/r1=2
As rate constant, k∝r
From Arrnhenius equation, we know that
log k2/k1= -Ea______ [T1-T2/T1T2]
log 2= -Ea______ [293-308/293x308]
0.3010= -Ea______ [-15/293x308]
=34673.48 J mol-1=34.7 kJ mol-1
The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R=8.314 J K-1 mol -1, log =0.6021]
We know, log
Given : T1 = 293 K, T2 = 313 K,
Let = K(consider) = 4K,
R = 8.314, Ea = ?
= 52854.43 = 52.8 K J mol-1
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume :
SO2Cl2(g) SO2 (g) + Cl2(g)
Experiment time/s-1 Total pressure/atm
1 0 0.4
2 100 0.7
Calculate the rate constant.
(Given :log 4 = 0.6021, log 2 = 0.3010)
SO2Cl2(g) SO2 (g) + Cl2(g)
Using formula, K =
When t=100 s
= = 1.386
Hydrogen peroxide, H2O2 (aq) decomposes to H2O(l) and O2(g) in a reaction that is first order in H2O2 and has a rate constant k = 1.0610-3 min-1.
(i) How long will it take for 15% of a sample of H2O2 to decompose?
(ii) How long will it take for 85% of a sample to decompose?
Given : k = 1.0610-3min-1
For first order reaction
k =where [ ]
t= 2172.64 log 1.1764
Time (t) =2172.64 0.07055
= 153.29 min
(ii) t= [for 85% completion]
t= 2172.64 log 6.66
Time (t) =2172.64 0.8234
= 1789.11 min