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प्रश्न
Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?
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उत्तर
Balanced chemical equation often leads to incorrect order or rate law. For example the following reaction seems to be a tenth order reaction.
\[\ce{KClO3 + 6FeSO4 + 3H2SO4 -> KCl + 3H2O + 3Fe2 (SO4)3}\]
This is actually a second order reaction. Actually the reaction is complex and occurs in several steps. The order of such reaction is determined by the slowest step in the reaction mechanism. Order is determined experimentally and is confined to the dependence of observed rate of reaction on the concentration of reactants.
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संबंधित प्रश्न
For a reaction: 
Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
For a reaction A + B ⟶ P, the rate is given by
Rate = k [A] [B]2
How is the rate of reaction affected if the concentration of B is doubled?
From the rate expression for the following reaction, determine the order of reaction and the dimension of the rate constant.
\[\ce{CH3CHO_{(g)} -> CH4_{(g)} + CO_{(g)}}\] Rate = k[CH3CHO]3/2
For the reaction:
\[\ce{2A + B -> A2B}\]
the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1. Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.
A reaction is first order in A and second order in B. How is the rate affected on increasing the concentration of B three times?
For a reaction R ---> P, half-life (t1/2) is observed to be independent of the initial concentration of reactants. What is the order of reaction?
Write resonating structures of ozone.
Rate law for the reaction \[\ce{A + 2B -> C}\] is found to be Rate = k [A][B]. Concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate constant will be ______.
Compounds ‘A’ and ‘B’ react according to the following chemical equation.
\[\ce{A(g) + 2B(g) -> 2C(g)}\]
Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
| Experiment | Initial concentration of [A]/mol L–¹ |
Initial concentration of [B]/mol L–¹ |
Initial rate of formation of [C]/mol L–¹ s–¹ |
| 1. | 0.30 | 0.30 | 0.10 |
| 2. | 0.30 | 0.60 | 0.40 |
| 3. | 0.60 | 0.30 | 0.20 |
In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction?
Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions?
Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II.
| Column I | Column II | |
| (i) |  |
|
| (ii) |  |
(a) 1st order |
| (iii) |  |
(b) Zero-order |
| (iv) |  |
Assertion: The enthalpy of reaction remains constant in the presence of a catalyst.
Reason: A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.
The role of a catalyst is to change
For a reaction A + B → products, the rate law is given by: r = `K[A]^(1/2)`. What is the order of reaction?
For a reaction \[\ce{Cl2l(g) + 2No(g) -> 2NaCl(g)}\] the rate law is expressed as rate= K[Cl2] [No]2 what is the order of the reaction?
For reaction 2A + B → BC + D which of the following does not Express the reaction rates
The half-life period of a. substance in a certain enzyme catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mol–1 to 0.04 mg L–1 is
A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is ______. (Use ln 2 = 0.693)
