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
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.
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;
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 
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