For a first-order reaction, what fraction of the starting material will remain a

Question

For a first-order reaction, what fraction of the starting material will remain after 4 half-lives? 1/16 1/9 1/4 1/8 1/3 A first-order reaction has a half-life of 1.10 s. If the initial concentration of reactant is 0.142 M, how long will it take for the reactant concentration to reach 0.00100 M? 0.127 s 4.50 s 7.87 s 1.59 s 3.09 s A first-order reaction has a rate constant of 0.021 s^-1. What is the half-life for this reaction? 0.69 s 0.015 s 14s 33 s 0.030 s The reaction 2NO_2(g) rightarrow 2NO(g) + O_2(g) is postulated to occur via the mechanism below: NO_2(g) + NO_2(g) rightarrow NO(g) + NO_3(g) k_1, slow NO_3(g) rightarrow NO(g) + O_2(g) k_2, fast The rate law for this mechanism is rate = k_2 [NO_3] rate = k_1 [NO_2] rate = k_1 [NO_2]^2 rate = k_1 k_2[NO_2]^2[NO_3] rate = k_1 k_2 [NO_3]^2

Explanation / Answer

1) Fraction remaing after n half lives = 1/(2n)

Thus, fraction remaining after 4 half lives = 1/24 = 1/16

2) The governing equation for 1st order reaction is :-

k*t = ln{[A0]/[A]} ; where k = rate constant , t = time , [A0] = initial concentration & [A] = concentration after time 't'

Also, rate constant, k = ln2/half life = 0.693/1.1 = 0.63 s-1

Thus, 0.63*t = ln(0.142/0.001)

or, t = 7.866 sec

3) rhalf life, t1/2 = ln2/rate constant = 0.693/0.021 = 33.007 s

4) The rate of a complex reaction, i.e a multiple step reaction is always govern by the slowest step of the reaction

Thus, the correct mechanism is :- (C) k1[NO2]2

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