Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
where Ea is the activation energy.
When a graph is plotted for a straight line with a slope of -4250 K is obtained. Calculate ‘Ea’ for the reaction.
(R = 8.314 JK-1 mol-1)
Ea --> Activation energy
The above equation is like y = mx + c where if we plot y v/s x we get a straight line with slope ‘m’ and intercept ‘c’.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:
t/s |
0 |
30 |
60 |
[CH3COOCH3] / mol L–1 |
0.60 |
0.30 |
0.15 |
(i) Show that it follows pseudo-first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time intervals 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
For the hydrolysis of methyl acetate to be a pseudo first-order reaction, the reaction should be first order with respect to ester when [H2O] is constant. The rate constant k for a first order reaction is given by:
Where,
[R]0 = intial concentration of reactant
[R] =final concentration of reactant
At =30s
We have
:
It can be seen that the rate constant k for the reaction has a constant value under any given time interval .Hence the given reaction follows pseudo-first order kinetics.
(ii) Average rate of reaction between the time interval 30-60 seconds is given by
Average rate =
Rate law for the reaction, A+ B → product is rate = k[A]2[B] What is the rate constant; if rate of reaction at a given temperature is 0.22 Ms-1, when [A]= 1 M band [BJ= 0.25 M?
3.52 M-2s-1
0.88 M-2s-1
1.136 M-2s-1
0.05 M-2s-1
B.
0.88 M-2s-1
The bacterial growth follows the rate law, dN/dt = kN where k is a constant and 'N' is the number of bacteria cell at any time. If the population of bacteria (number of cell) is doubled in 5 min find the time in which the population will be eight times of the initial one?
As dN/dt = kN (It is a Ist order reaction)
Integrated rate equation for Ist order reaction
k = 2.303/t log N/No
when t = 5 min, N = 2No
k = 2.303/5 log 2No/No
or k = 2.303/5 log 2
for N = 8No , t = ?
t = 2.303/k log 8No/No
By putting the value of k
t = log 8No/N
or t = = 15 min
Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:
6.93×10−4 mol min−1
2.66 L min−1 at STP
1.34×10−2 mol min−1
1.34×10−2 mol min−1
A.
6.93×10−4 mol min−1
For first order reaction