For a certain reaction, rate = k × [H⁺]ⁿ. If pH of reaction changes from two to one, the rate becomes 100 times of its value at pH = 2, the order of reaction is –

1
2
zero
3

Solution

Activation energy of a chemical reaction can be determined by

evaluating rate constant at standard temperature
evaluating velocities of reaction at two different temperatures
evaluating rate constants at two different temperatures
changing concentration of reactants

Solution

Consider the following statements:
I.Increase in concentration of reactant increases the rate of a zero order reaction.
II.Rate constant k is equal to collision frequency A if E_a = 0
III.Rate constant k is equal to collision frequency A if E_a= ∞.
IV.In k vs T is a straight line.
V.In k vs 1/T is a straight line.
Correct statements are

I and IV
II and V
III and IV
II and III

Solution

A chemical reaction is catalyzed by a catalyst X. Hence X

reduces enthalpy of the reaction
decreases rate constant of the reaction
increases activation energy of the reaction
does not affect equilibrium constant of the reaction

Solution

A catalyst affects equally both forward and backward reactions, therefore it does not affect equilibrium constant of reaction.

For an exothermic reaction, the energy of activation of the reactants is

equal to the energy of activation of products
less than the energy of activation of products
greater than the energy of activation of products
sometimes greater and sometimes less than that of the products

Solution

Activation energy of reactant is less than the energy of activation of products.

The rate law for the reaction 2X + Y ⟶ Z is Rate =k[X][Y].The correct statement with regard to this relation is

the unit of k is s^–1
the rate of the reaction is independent of [X] and [Y]
for this reaction t_1/2 is independent of initial concentrations of reactant
the rate of formation of Z is twice the rate of disappearance of X

Solution

What is the half life of a radioactive substance if 87.5% of any given amount of the substance disintegrates in 40 minutes ?

20 minutes
10 minutes
13 minutes 32 sec
160 minutes

Solution

t_{\(\frac{1}{4}\)} can be taken as the time taken for the concentration of a reactant to drop to \(\frac{3}{4}\) of its initial value.If the rate constant for a first order reaction is k, the t_{\(\frac{1}{4}\)} can be written as

0.75/k
0.69/k
0.29/k
0.10/k

Solution

Ethylene is produced by \(C_{4}H_{8} \overset{\Delta }{\rightarrow} 2C_{2}H_{4}\) The rate constant is 2.3 × 10^–4 sec^–1. Approximately in what time will the molar ratio of ethylene to cyclobutane in mixture attain the value equal to one ?(log 2 =0.3, log 3 = 0.47)

1700 sec
1600 sec
2000 sec
1200 sec

Solution

The half-life of a reaction is inversely proportional to the square of the initial concentration of the reactant. Then the order of the reaction is

zero
1
2
3

Solution

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