1. Given reaction

$A(g) \rightarrow 2B(g)+C(g)$

This is a first-order decomposition.

2. Relate total pressure to extent of reaction

Let initial pressure of pure $A$ be

$P_0=800\ \text{mm Hg}$

Suppose after time $t$, fraction decomposed is $\alpha$.

Then:

• moles of $A$ left $=1-\alpha$
• moles of $B$ formed $=2\alpha$
• moles of $C$ formed $=\alpha$

So total moles at time $t$ are

$n_t=(1-\alpha)+2\alpha+\alpha=1+2\alpha$

Since pressure is proportional to total moles at constant $T,V$,

$P_t=P_0(1+2\alpha)$

3. Use data at 10 min

At $t=10$ min, total pressure is 1600 mm Hg:

$1600=800(1+2\alpha)$

$2=1+2\alpha$

$2\alpha=1$

$\alpha=\frac{1}{2}$

So after 10 min, 50% of $A$ has decomposed.

4. Use first-order kinetics

For a first-order reaction,

$[A]_t=[A]_0 e^{-kt}$

and since fraction unreacted is $1-\alpha$,

$1-\alpha=e^{-kt}$

At $t=10$ min, $\alpha=\frac12$, so

$e^{-10k}=\frac12$

Thus,

$k=\frac{\ln 2}{10}$

5. Find decomposition after 30 min

At $t=30$ min,

$e^{-30k}=e^{-3\ln 2}=\frac{1}{8}$

So fraction of $A$ left is $\frac18$, hence fraction decomposed is

$\alpha=1-\frac18=\frac78$

6. Calculate total pressure after 30 min

$P_{30}=800\left(1+2\times \frac78\right)$

$P_{30}=800\left(1+\frac{14}{8}\right)=800\left(\frac{11}{4}\right)$

$P_{30}=2200\ \text{mm Hg}$

7. Final answer

$\boxed{2200}$