Determine the average life of U 238 having t 1 2 = 140 days .

Given, t 1 2 = 140 days Therefore, radioactive element U 238 , follows the first order reaction. So, according to first order reaction- t 1 2 = ln 2 K = 0 . 69 K - - - - 1 t average = 1 k - - - - 2 From equation (1) and (2) t 1 2 = 0 . 69 × t average 140 = 0 . 69 × t average t average = 140 0 . 69 = 203 days Therefore, the average life of U 238 will be 203 days (approx).

Determine the average life of U 238 having t 1 2 = 140 days .

Given, t 1 2 = 140 days Therefore, radioactive element U 238 , follows the first order reaction. So, according to first order reaction- t 1 2 = ln 2 K = 0 . 69 K - - - - 1 t average = 1 k - - - - 2 From equation (1) and (2) t 1 2 = 0 . 69 × t average 140 = 0 . 69 × t average t average = 140 0 . 69 = 203 days Therefore, the average life of U 238 will be 203 days (approx).

Write any two characteristics of first order reaction.

In first order reactions, the rate of reaction is directly proportional to the first power of the concentration of the reactant. Integrated rate equation- k = 2 . 303 t log A ∘ A For example- Decomposition of H 2 O 2 , H 2 O 2 → H 2 O + 1 2 O 2 Following are the characteristics of first order reaction- Concentration of reactants is totally independent of rate constant. Straight line is formed between rate versus concentration graph. Half life period is constant.

Explain the experimental determination of the rate constant for the decomposition of H 2 O 2 in an aqueous solution.

The decomposition of hydrogen peroxide in an aqueous solution is catalysed by iodide ions. This reaction completes in two steps: ( i ) H 2 O 2 aq + I - → H 2 O aq + OI - aq ii OI - aq + H 2 O 2 aq → H 2 O aq + I - aq + O 2 It is clear that the slowest step of the reaction is the rate-determining step. In this given reaction, the first step is the slowest and rate-determining step. Hence, the rate of the reaction = k H 2 O I - So, the rate of the reaction will be 1 .

The activation energy of a certain reaction is 100 kJ / mole . What is the change in the rate constant of the reaction if the temperature is changed from 25 ∘ C to 35 ∘ C ? Let the rate constants at 25 ∘ C be K 1 and at 35 ∘ C be K 2 respectively.

Given, E a = 100 KJ / mole = 100 × 10 3 J / mol K 1 = 25 ∘ C = 298 K K 2 = 35 ∘ C = 308 K We know that, log K 2 K 1 = E a 2 . 303 R 1 T 1 - 1 T 2 log K 2 K 1 = 100 × 10 3 2 . 303 × 8 . 314 1 298 - 1 308 log K 2 K 1 = 100 × 10 3 2 . 303 × 8 . 314 308 - 298 308 × 298 log K 2 K 1 = 10 6 1 . 75 × 10 6 K 2 K 1 = 3 . 72 ∴ K 2 ≈ 4 K 1

Calculate the number of α and β particles emitted in the conversion of 90 Th 232 to 82 Pb 208 .

Let a be the α -particles and b be the β -particles emitted in the conversion of 90 Th 232 to 82 Pb 208 . 90 Th 232 → 82 Pb 208 + a He 2 4 + b e - 1 0 ∴ No . of α particles = 232 - 208 4 = 24 4 = 6 and, No . of β particles = 2 × n - reactant - product = 2 × 6 - 90 - 82 = 4 Therefore, the number of α and β particles emitted in the conversion of 90 Th 232 to 82 Pb 208 is 6 and 4 respectively.

Consider the potential energy diagrams of reactions (I) and (II) given below, predict which reaction will go faster and why ?

The excess energy which should be supplied to the reactants to undergo chemical reactions is called activation energy, E a . It is equal to the difference between the threshold energy needed for the reaction and the average kinetic energy of all the reacting molecules. We already know that, Activation energy = Threshold energy - Average energy of the reacting molecules E a = E threshold - E reactants If the activation energies are low, then the reaction will be fast. Whereas, slow reactions is the result of high activation energies. Therefore, (II) will have the faster reaction than (I) because (II) has low activation energy.

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Determine the average life of $U^{238}$ having $t_{\frac{1}{2}} = 140 days$ .

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