Chemical Kinetics and Nuclear ChemistryHard

Question

Decomposition of A2(g) and B3(g) follows first order kinetics as follows.

$${A_{2}(g) \rightarrow 2A(g);K_{1}\left( /h^{- 1} \right) = e^{- \frac{14000(/J)}{RT} + 5} }{B_{3}(g) \rightarrow 3B(g);K_{2}\left( /h^{- 1} \right) = e^{- \frac{20000(/J)}{RT} + 10}}$$

Here, K1 and K2 are the rate constants with respect to disappearance of A2 and B3, respectively. One mole, each of A2(g) and B3(g), is taken in a 100 L evacuated flask and at some temperature at which they start decomposing at the same rate. The incorrect information regarding the reactions is

Options

A.At any instant, $\left( \frac{P_{A}}{P_{B}} \right)$ will be constant and equal to 1.0, where PA and PB are the partial pressures of gases at that time.
B.At any instant, $\left( \frac{P_{A}}{P_{B}} \right)$ will be constant and equal to 1.0, where PA and PB are the partial pressures of gases at that time.
C.At some instant, the total pressure of all gases in the flask may be less than 0.2 atm.
D.At any instant, $\left( \frac{P_{A}}{P_{B}} \right)$ will be constant and equal to 1.0, where PA and PB are the partial pressures of gases at that time.

Solution

$K_{1} = K_{2} \Rightarrow - \frac{14000}{RT} + 5 = - \frac{20000}{RT} + 10$

$\Rightarrow T = \frac{1200}{8.314}\text{ K}$

Now, $\frac{P_{A_{2}}}{P_{B_{3}}} = \frac{1 \times e^{- K_{1}t}}{1 \times e^{- K_{2}t}} = \frac{1}{1}$

Now, initial pressure

$P_{0} = \frac{(1 + 1) \times 0.0821 \times \frac{1200}{8.314}}{100} = 0.237\text{ atm}$

As number of moles will increase on reaction, the total pressure can never be less than 0.2 atm

Now, $\frac{P_{A}}{P_{B}} = \frac{2K_{1}}{3K_{2}} = \frac{2}{3}$

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