Write the equation relating the half-life ( t 1 2 ) of a reaction and the rate constant ‘k for first order reaction.

For the first order reaction, k = 2 . 303 t log [ R ] 0 [ R ] at t 1 2 [ R ] = [ R ] 0 2 So , the above equation becomes k = 2 . 303 t log [ R ] 0 [ R ] 0 2 or t 1 2 = 2 . 303 k log 2 t 1 2 = 2 . 303 k × 0 . 301 t 1 2 = 0 . 693 k It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa.

Write the equation useful to calculate half-life ( t 1 2 ) values for zero and first order reactions.

For a zero order reaction, rate constant is given by equation k = [ R ] 0 - [ R ] t At t = t 1 2 , [ R ] = 1 2 [ R ] 0 The rate constant at t 1 2 becomes k = [ R ] 0 - 1 2 [ R ] 0 t 1 2 t 1 2 = [ R ] 0 2 k It is clear that t 1 2 for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant. For the first order reaction, k = 2 . 303 t log [ R ] 0 [ R ] at t 1 2 [ R ] = [ R ] 0 2 So , the above equation becomes k = 2 . 303 t log [ R ] 0 [ R ] 0 2 or t 1 2 = 2 . 303 k log 2 t 1 2 = 2 . 303 k × 0 . 301 t 1 2 = 0 . 693 k

State and explain kohlrausch's law of independent migration of ions.

Weak electrolytes like acetic acid have lower degree of dissociation at higher concentrations and hence for such electrolytes, the change in Λ m with dilution due to the increase in the degree of dissociation and consequently the number of ions in total volume of solution that contains 1 mole of electrolyte. In such cases, Λ m increases steeply on dilution, especially near lower concentrations. Therefore, Λ m o cannot be obtained by the extrapolation of Λ m to zero concentration. At infinite dilution, electrolyte dissociates completely but at such low concentration, the conductivity of the solution is so low, that it cannot be measured experimentally. In such cases, Λ m o for weak electrolytes can be obtained by using kohlrausch's law. Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte. Thus, Λ m o = ν + λ + o + ν - λ - o Here, λ + o , λ - o are the limiting molar conductivities of the cation and anion respectively.

What are galvanic cells? Explain the working of a galvanic cell with a neat sketch taking Daniel cell as example.

Galvanic cell: A device which converts chemical energy into electrical energy by the use of spontaneous redox reactions is called Galvanic cell or voltaic cell. E.g: Daniel cell. Daniel cell: It is a special type of galvanic cell. It contains two half cells in the same vessel. The vessel is divided into two chambers. Left chambers is filled with ZnSO 4 ( aq ) solution and Zn-rod is dipped into it. Right chamber is filled with CuSO 4 ( aq ) solution and copper rod is dipped into it. Process diaphragm acts as a salt bridge. The two half cells are connected to external battery. Cell reactions: Ion Zn / ZnSO 4 half cell, oxidation reaction occurs Zn → Zn 2 + + 2 e - Ion Cu / CuSO 4 half cell, reduction reaction occurs. Cu 2 + + 2 e - → Cu The net cell reaction is Zn + Cu 2 + ⇔ Zn 2 + + Cu Cell is represented as Zn / Zn 2 + ∥ Cu 2 + / Cu

Write the Arrhenius equation for the rate constant (k) of a reaction.

The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation. k = A e - Ea RT where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and E a is activation energy measured in joules/mole ( J mol - 1 ) .

What are pseudo first order reactions? Give one example.

The reactions which are not truly of first order but become reactions of first order under certain conditions are called pseudo first order reactions. For example, hydrolysis of ester. Rate = k ' [ CH 3 COOC 2 H 5 ] [ H 2 O ] Since water is in excess and [ H 2 O ] is constant . So , Rate = k [ CH 3 COOC 2 H 5 ] , where k = k ' [ H 2 O ]

What are elementary reactions?

Elementary reactions: A balanced chemical equation never gives us a true picture of how a reaction takes place since rarely a reaction gets completed in one step. The reactions taking place in one step are called elementary reactions or an elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state.

The rate constant for a first order reaction is 60 s - 1 . How much time will it take to reduce the initial concentration of the reactant to its 1 / 16 th value?

Given, rate constant k for a first order reaction = 60 s - 1 Let us assume: Initial concentration R 0 = x Final concentration R = x 16 Formula for first order reaction, t = 2.303 k l o g R 0 R = 2.303 k l o g x x 16 = 2.303 k l o g 1 1 16 = 2.303 60 l o g 16 = 4.6 × 10 - 2 s Hence, the required time is 4.6 × 10 - 2 s .

A first order reaction takes 40 minutes for 30 % decomposition. Calculate t 1 / 2 ?

Given data: Order of reaction = 1 R 0 = 100 ; R = 100 - 30 = 70 First order rate equation is: k = 2 .303 t log R 0 R k = 2.303 40 l o g 100 70 = 2.303 40 l o g 10 7 = 8.91 × 10 - 3 m i n - 1 Therefore, t 1 / 2 of the decomposition reaction can be determined as follows: t 1 / 2 = 0 .693 k = 0.693 8.91 × 10 - 3 = 77.71 m i n .

Derive an integrated rate equation for a first order reaction.

Let us, consider a unimolecular first order reaction represented by the general equation, A → Product At t = 0 a 0 At t = t a - x x The initial concentration of A is a mole L - 1 and its concentration after time t is ( a – x ) mole L - 1 . This means during the time interval t , x mole L – 1 of A has reacted. The rate of reaction at any time t is given by the following first order kinetics. dx dt = k a - x where k is the rate constant of the reaction. dx a - x = kdt This is differential rate equation and can be solved by integration. ∫ dx a - x = k ∫ dt - ln a - x = k . t + C . . . . . ( 1 ) where C is integration constant. The constant C can be evaluated by applying the initial condition of the reaction i.e. when t = 0 , x = 0 . Putting these in equation (1), we get C = – ln a Putting the value of C in equation (1), we get – ln ( a – x ) = k . t – lna k = 1 t ln a a - x = 2 . 303 t log a a - x . . . . . ( 2 ) Equation (2) integrated rate equation for a first order reaction.