Question

The isomerization reaction ${\text{CH}}_{\text{3}}\text{NC}\to {\text{CH}}_3\text{CN}$ obeys the first-order rate law in the presence of an excess of argon. Measurements at 500 K reveal that in 520 s the concentration of CH₃NC decreases to 71% of its original value. Calculate the rate constant k of the reaction at 500 K.

Short answer

Rate constant value at 500K is$6.6\times {10}^{-4}{s}^{-1}$

Step-by-step solution

Rate expression:

The eqaution for the first order reaction when initail and final concendration are given

$$\begin{gathered} \ln \left(\frac{{\text{C}}_0}{{\text{C}}_t}\right)=kt.......\left(1\right) \\ or \\ 2.303\log \left(\frac{{\text{C}}_0}{{\text{C}}_t}\right)\hspace{0.17em}=kt.......\left(2\right) \end{gathered}$$

Where C₀ and C_t are initial and final concendration ,k is the rate constant and t is the time in seaconds.

: Calculation of  Rate constant K value:  

As the concendration is decreased in terms of percentage the final concendration can be written as

$$\begin{gathered} {\text{C}}_t\text{=}\frac{C_0\times 71}{100}\text{.............................}\left(3\right) \\ \text{Substitutingallthesevaluesinequation2} \\ 2.303\log \left(\frac{C_0}{{\displaystyle \frac{C_0\times 71}{100}}}\right)=\text{k}\times 520s \\ k=\frac{2.303}{520}\log \left(\frac{100}{71}\right)s^{-1} \\ k=\hspace{0.17em}6.6\times {10}^{-4}{s}^{-1} \end{gathered}$$

Hence the rate constant value is$6.6\times {10}^{-4}{s}^{-1}$