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(2)

$K_{1} = \frac{\ln 2}{12}, K_{2} = \frac{\ln 2}{3}$

$K = K_{1} + K_{2} = \frac{\ln 2}{12} + \frac{\ln 2}{3}$

$\frac{\ln 2}{t} = \frac{\ln 2}{12} + \frac{\ln 2}{3} \Rightarrow \frac{1}{t} = \frac{1}{12} + \frac{1}{3}$

$\frac{1}{t} = \frac{5}{12} \Rightarrow t = \frac{12}{5} = 2 min$

(3)

$a_{t} = a_{0} \times \frac{12.5}{100} = \frac{a_{0}}{8} (87.5 % completion)$

$a_{0} \overset{t_{1 / 2}}{\to} a_{0 / 2} \overset{t_{1 / 2}}{\to} a_{0 / 4} \overset{t_{1 / 2}}{\to} \frac{a_{0}}{8}$

$t_{87.5} = 3 t_{1 / 2}$ $∴$ $x = 3$

(2200)

$\begin{matrix} A (g) & ⟶ & 2 B (g) & + & C (g) \\ t = 0800 & - & - \\ t = (10 min) & 800 - P & 2 P & P \end{matrix}$

$P_{T} = 800 - P + 2 P + P$

$= 800 + 2 P = 1600$

So $P = 400 mm$

So $t_{1 / 2} = 10 min$

Given time = 30 min.

So for 1st order reaction $C_{t} = \frac{C_{0}}{{(2)}^{n}}$

$n = \frac{t}{t_{1 / 2}} = \frac{30}{10} = 3$ ; $n =$ number of half life

$C_{t} = \frac{800}{{(2)}^{3}} = 100 mm of Hg$

So After 30 minutes, $P = 700 mm$

$P_{A} = 800 - 700 = 100,$ $P_{B} = 1400,$ $P_{C} = 700$

$P_{T}$ of reaction mixture $= 100 + 1400 + 700 = 2200 mm of Hg$

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