Consider the following single step reaction in gas phase at constant temperature.

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Solution

Q.
Consider the following single step reaction in gas phase at constant temperature.

$2 A_{(g)} + B_{(g)} \to C_{(g)} $

The initial rate of the reaction is recorded as $r_{1}$ when the reaction starts with 1.5 atm pressure of A and 0.7 atm pressure of B. After some time, the rate $r_{2}$ is recorded when the pressure of C becomes 0.5 atm. The ratio $r_{1} : r_{2}$ is _________ $\times 10^{- 1} .$ (Nearest Integer) [2024]

Ans.
(315)

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

t = 0 1.5 atm 0.7 atm

(press)

t = t (1.5 - 2p) atm (0.7 - p) atm p atm

(press)

  Since reaction is single step reaction, its rate law can be written from stoichiometry of the reaction as:

   $r = k {(P_{A})}^{2} (P_{B})$

Initial rate $r_{1}$ is thus $r_{1}$ = $k {(1.5)}^{2} (0.7)$

As given in question p = 0.5 atm

Thus pressure of A after sometime

   $= (1.5 - 2 p) atm = (1.5 - 2 \times 0.5) atm = 0.5 atm$

Pressure of B after sometime

   $= (0.7 - p) atm = (0.7 - 0.5) atm = 0.2 atm$

  Rate after sometime, $r_{2} = k {(0.5)}^{2} (0.2)$

$\frac{r_{1}}{r_{2}} = \frac{k {(1.5)}^{2} (0.7)}{k {(0.5)}^{2} (0.2)} = 315 \times 10^{- 1}$

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