Home work for chemical reaction kinetics 1.1t. For a second-order reaction, 1/3

Question

Home work for chemical reaction kinetics 1.1t. For a second-order reaction, 1/3 of the reactant is spent in 10 minutes, how mach time should be taken for reaction of another 1/3 reactant? (C) 30 ID) 40 min (A) 10min (B) 20 min min . 1.12 In the collision theory of bimolecular gas-phase reaction, we can say that A) Every collision of the activated reactant molecules will lead to reaction and relevant products. (B) A "head-on collision of reactant molecules will definitely lead to reaction (C) Only head-on, collision of activated molecules will lead to reaction. (D) A reacting molecular pair should have a colliding relative kinetic energy, with a component not less than a threshold value on the axis connecting the center of the molecular pair 1.13. if a reversible reaction has a heat of reaction of-50 kJ mol at constant volume, the activation energy of the reverse reaction (A) must be equal or less than 50 kJ mor (B) must be equal or greater than 50kJ mol (C) can be less or greater than 50 kJ mol (D) must be less than 50 kJ mo 1.14. For a surface-reaction-controlled gas-solid catalytic reaction A ? B, the rate equation is rkbaPA/+bP) If the adsorption of the reactant is strong enough at a moderate pressure, the reaction should behave as) (A) a zero-arder reaction 18) a first-order reaction IC) a second-order reaction(Dja third-order reaction 1.15. A gas-phase unimolecular reaction can be elucidated by the mechanism The deduced rate equation is Therefore the reaction should be A) first-order at high pressures and second-order at low pressures (B) second-order at high pressures and first-order at low pressures (C) first-order at high pressures and zero-order at low pressures (D) second-order at high pressures and zero-order at low pressures. 1.16. If a reaction conforms to the kinetic equation k (Vt) (Co-C)/CoC)l, the reaction ( A) must be a unimolecular reaction.

Explanation / Answer

Ans 1.11

For second order reaction

1/[A] - 1/[A0] = k*t

[A0] = Initial concentration of reactant A at time t = 0

[A] = [A0]/3 = final concentration of reactant A at time t = 10 min

1/([A0]/3) - 1/[A0] = k*10

k = 0.2/[A0]

Again

Now initial concentration = [A0]/3

Final concentration = [A0]/9

Rate constant k = 0.2/[A0]

1/ ([A0]/9) - 1/([A0]/3) = t*0.2/[A0]

6/[A0] = 0.2t/[A0]

6 = 0.2t

t = 30 min

Option C is the correct answer

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