Step 1. Since $N_2$ gas evolved is proportional to the amount of $C_6H_5N_2Cl$ REACTED, $V_\infty$ (total gas at complete reaction) is proportional to the initial concentration $[A]_0$, and $(V_\infty-V_t)$ is proportional to the concentration REMAINING at time t. So the first order formula becomes $k=\dfrac{2.303}{t}\log\dfrac{V_\infty}{V_\infty-V_t}$.

Step 2. At $t=6$: $V_\infty-V_t=58.3-19.3=39.0$; $k=\dfrac{2.303}{6}\log\dfrac{58.3}{39.0}=0.3838\times0.1746=0.0670\ \text{min}^{-1}$.

Step 3. At $t=12$: $58.3-32.6=25.7$; $k=\dfrac{2.303}{12}\log\dfrac{58.3}{25.7}=0.1919\times0.3557=0.0683\ \text{min}^{-1}$.

Step 4. At $t=18$: $58.3-41.3=17.0$; $k=\dfrac{2.303}{18}\log\dfrac{58.3}{17.0}=0.1279\times0.5352=0.0685\ \text{min}^{-1}$.

Step 5. At $t=24$: $58.3-46.5=11.8$; $k=\dfrac{2.303}{24}\log\dfrac{58.3}{11.8}=0.0960\times0.6938=0.0666\ \text{min}^{-1}$. …

Treat $V_\infty$ as proportional to initial concentration and $(V_\infty-V_t)$ as proportional to remaining concentration, compute k at each time point …

• Using $V_t$ directly (instead of $V_\infty-V_t$) as the 'remaining concentration' -- here $V_t$ tracks gas EVOLVED (i.e. product formed / amount REACTED), so the remaining reactant is proportional to $V_\infty-V_t$, …