Decomposition of reaction $3A\left(g\right)⟶2B\left(g\right)+2C\left(s\right)$ follows first order kinetics. Starting with pure A (at 6 atm), the pressure developed after 20 minutes and after a long time are 5.05 atm and 4.05 atm, respectively. Identify the correct statement.

$$\begin{gathered} 3A\left(g\right) ⟶ 2B\left(g\right) + 2C\left(s\right) \\ t=0 6 - - \\ \begin{array}{ccc}t=t& 6-x& \frac{2x}{3} - \\ t=\mathrm{\infty }& 0 & - -\end{array} \end{gathered}$$

Final pressure should be 4 atm, however, given is 4.05 atm. This means some C might have volatilized and its vapour pressure = 4.05 - 4 = 0.05

Now, At  $6-x+\frac{2x}{3}=5.05-0.05$

$\begin{array}{l}6-\frac{x}{3}=5\\ \Rightarrow x\mathit{ }\mathit{ }\mathit{ }\mathit{ }=3\\ \Rightarrow t_{1/2}\mathit{ }\mathit{ }\mathit{ }\mathit{ }=20\min \\ \therefore t_{75\mathrm{\%}}=40;t_{87.5\mathrm{\%}}=60\\ t_{9375}=80\min \end{array}$