The isomerization reaction CH3NC CH3CN obeys the first-order rate law, rate = k[

Question

The isomerization reaction CH3NC CH3CN obeys the first-order rate law, rate = k[CH3NC], in the presence of an excess of argon. Measurement at 500. K reveals that in 487 seconds, the concentration of CH3NC has decreased to 73% of its original value. Calculate the rate constant (k) of the reaction at 500. K. s1 (The integrated form for the first-order rate law can be written in the general terms ln[A]t ln[A]0 = kt, where [A]0 is the initial concentration of reactant A, [A]t is the concentration of A at time t, and k is the rate constant.)

Explanation / Answer

we have:
[CH3NC]o = 100 M (Let initial concentration be 100)
[CH3NC] = 73 M (73 % is remaining)
t = 487 s


use integrated rate law for 1st order reaction
ln[CH3NC] = ln[CH3NC]o - k*t
ln(73) = ln(100) - k*487
4.2905 = 4.6052 - k*487
k*487 = 0.3147
k = 6.46*10^-4 s-1
Answer: 6.46*10^-4 s-1

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