A gaseous phase reaction A₂→ B + 0.5C shows an increase in pressure from 100 mm to 120 mm in 5 min. Now, $-\frac{\mathrm{\Delta }\left[{\mathrm{A}}_2\right]}{\mathrm{\Delta t}}$should be

$$\begin{gathered} {\mathrm{A}}_2 \to \mathrm{B} + 0.5\mathrm{C} \\ 100 0 0 \\ 100-\mathrm{p} \mathrm{p} 0.5\mathrm{p} \\ 100-\mathrm{p}+\mathrm{p}+0.5\mathrm{p}=120 \\ \mathrm{p}=\frac{20}{0.5} =40 \\ \mathrm{so} ∆{\mathrm{A}}_2=100-60=40 \\ -\frac{∆{\mathrm{A}}_2}{∆\mathrm{t}} =\frac{40}{5}=8\mathrm{mm} - {\min }^{-1} \end{gathered}$$