For a reaction :
(i)Write the order and molecularity of this reaction.
(ii)Write the unit of k.
(i) This reaction is catalysed by Pt at high pressure. So, it is a zero-order reaction with molecularity 2.
(ii) The rate law expression for this reaction is Rate = k
Hence, the unit of k is mol L−1 s−1.
A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Let the concentration of the reactant be [A] = a
Rate of reaction, R = k [A]2 = ka2
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k (2a)2
= 4ka
= 4R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e., then the rate of the reaction would be
Therefore, the rate of the reaction would be reduced to
The reaction N2(g) + O2(g) 2NO(g), contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 x 10-5. Suppose in a case [N2] = 0.80 mol L-1and [O2] = 0.20 mol L-1 before any reaction occurs. Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.
N2(g) + O2(g) 2NO(g)
T=0 0.8 0.20
T=t 0.8-x 0.2-x 2x
Kc = 1.0 x 10-5
Kc =
Or, 1.0 x 10-5 = (2x)2/ [(0.8 – x)(0.2 – x )]
If x is very small, then
0.8 – x 0.8
0.2 – x 0.2
1.0 x 10-5 = (2x)2 / [(0.80) (0.2)]
16 x (10-6) = 4 x2
X2 = 2 x 10-3
Therefore, the amount of reactant and product at equilibrium is as follows:
N2 = 0.8 - 0.002 = 0.798
O2 = 0.2 - 0.002 = 0.198
NO = 2x = 2 x 2 x 10-3 = 4 x 10-3
The rate constant for the first-order decomposition of H2O2 is given by the following equation:
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes. (Given: R = 8.314 JK–1mol–1).
Given:
Order of the reaction = First order
t1/2 = 200 minutes = 200 × 60 = 12,000 seconds The relation between t1/2
and k is given by t1/2 = 0.693/k
k = 0.693/12000 = 5.7 × 10−5
The rate constant for the first-order decomposition of H2O2 is given by
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 4 = 0·6021]
Given: T1 = 293 k
T2 = 313 k
R = 8.314 J k-1mol-1
k2 = 4k1
Ea =?
The formula used is;