Question

The isomerization reaction ${\mathbf{\text{CH}}}_{\mathbf{3}}\mathbf{\text{NC}}\mathbf{}\mathbf{\to }{\mathbf{\text{CH}}}_{\mathbf{3}}\mathbf{\text{CN}}$ obeys the first-order rate law in$\mathbf{\text{Ratelaw}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\text{-k}}\left[{\text{CH}}_3\text{NC}\right]$ 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$6.6\times {10}^{-4}{\text{s}}^{-1}$.

Step-by-step solution

Step-1: Rate expression:

The equation for the first order reaction when initial and final concentration are given

$\begin{array}{c}\ln \left(\frac{C_0}{Ct}\right)=kt........\left(1\right)\\ or\\ 2.303\log \left(\frac{C_0}{C_t}\right)=kt..........\left(2\right)\end{array}$

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

Step-2: Calculation of  Rate constant K value:  

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

$\begin{array}{c}{C}_t=\frac{C_0\times 71}{100}.......\left(2\right)\\ \text{Substitutingallthesevaluesinequation2}\\ 2.303\log \left(\frac{C_0}{\frac{C_0\times 71}{100}.}\right)=k\times 520s\\ k=\frac{2.303}{520}\log \left(\frac{100}{71}\right)s^{-1}\\ k=6.6\times {10}^{-4}{s}^{-1}\end{array}$

Hence rate constant value at 500K is$6.6\times {10}^{-4}{\text{s}}^{-1}$.