Submit The second order reaction A → Products takes 13.5 s for the concentration of A...
The second order reaction A → Products takes 13.5 s for the concentration of A to decrease from 0.740 M to 0.285 M. What is the value of k for this reaction?
It takes 50.5 s for the concentration of reactant A in the second order reaction A ==> Products 0.84 mol L to half of it. to decrease from its initial value [A]o A) What is the rate constant of the reaction? B) What is the concentration of A after 32 s have passed? B) After what time will the concentration of A be [A]o 16?
The half-life of a reaction, t1/2, is the time it takes for the reactant concentration [A] to decrease by half. For example, after one half-life the concentration falls from the initial concentration [A]0 to [A]0/2, after a second half-life to [A]0/4, after a third half-life to [A]0/8, and so on. on. For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as t1/2=0.693k For a...
The rate constant for this second-order reaction is 0.430 M-'.s at 300 °C. A- products How long, in seconds, would it take for the concentration of A to decrease from 0.670 M to 0.310 M? 1 = 6.355 Incorrect Calculate the rate constant, k, for a reaction at 56,0 °C that has an activation energy of 88.6 kJ/mol and a frequency factor of 6.85 x 10's-1 k= 2.2188 SI Incorrect
A reaction A B has the following time dependence for the concentration of [A] vs time. For t=(0 s, 5 s, 10 s, 15 s, 25 s) the concentration of [A]=(30.00 M, 19.81 M, 13.08 M, 8.64 M, 3.77 M). The initial concentration of [A] is the value at t=0 s. (A)Calculate the values of the rate constant k assuming that the reaction is first order. What is the value of k at 5 s? s-1 What is the value...
2. The reaction A → products was found to be second order order and have a rate constant, k, of 0.681 M-1 5-1. If the initial concentration of the reaction was 0.885 M, what is the half life for the reaction? 10.2 Submit Answer Incorrect. Tries 5/45 Previous Tries
7. Consider the following second-order reaction: 2X ? Y. The initial concentration of X is 0.50 M. It takes 30 minutes for the concentration of X to decrease to 35% of its initial value. Calculate the rate constant for this second-order reaction. A. 0.0330 Mmin B. 0.124 M-min-1 C. 0.185 M-'min-1 D.0.367 Mmin-1 E. 2.48 M-min-1
[time] [Z] 4. Suppose the data shown are for the second order reaction: → products. What is the value of the rate constant k? O min 50.000 atm 6.250 3.333 2.273. 5. For a first order reaction (G - products) with k = 0.173 min', suppose a chemist runs the reaction starting with an initial concentration [G]. = 12.0 M. a. How many minutes will it take for [G] to decrease to 4.70 M? b. What [G] will remain after...
the rate equation for the second order reaction of E and S is what os the diffusion controlled limit in aqueous solution sapling learning The rate equation for the second-order reaction of E and S is where kcat/KM is the second-order rate constant, which can be used to compare catalytic efficiencies cat What is the diffusion-controlled limit in aqueous solution? O 107 to 108 M s O 108 to 10 M-1s-1 O 10-6 to 10-5 M O 109 to 1010...
1. A reaction is second order in[A] and second-order in [B]: Rate,=K[A]^2[B]^2. what are the units of k for this reaction? If the concentration of air decreases by a factor of 2 and the concentration of b increases by a factor of 5 what happens to the rate? 2. for the forward reaction 2NO+Cl2=>2NOCl. determine the rate(m/s)for experiment #4 given [NO]°(M)=0.40M and [Cl2]°z(M)=0?20M. Rate? 3.The following data were collected over time for the forward reaction 2NO2=>2NO+O2 ( 1/[NO2]=100 at 0...