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Standard XII

Chemistry

Elevation in Boiling Point

The gaseous d...

Question

The gaseous decomposition reaction: $A (g) \to 2 B (g) + C (g)$ , is observed to first order over the excess of liquid water at $25^{o} C$ . It is found that after 10 minutes, the total pressure of the system is 188 torr and after a very long time it is 388 torr. Calculate the rate constant of the reaction in $h r^{- 1}$ . The vapour pressure of $H_{2} O$ at $25^{o} C$ is 28 torr. $[l n 2 = 0.7, l n 3 = 1.1, l n 10 = 2.3]$

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Solution

The correct option is B $1.2$
$A (g) \to 2 B (g) + C (g)$
Let the initial pressure be $P_{0}$ 0 0
After 10 mins, $(P_{0} - x)$ $2 x$ $x$
After long time $(t \to \infty)$ 0 $2 P_{0}$ $P_{0}$
Now,
$(P_{0} - x) + 2 x + x +$ vapour pressure of $H_{2} O = 188$ .
$P_{0} + 2 x = 160$ and $3 P_{0} + 28 = 388$
So, $P_{0} = 120$ and $x = 20 t o r r$ .
$k = \frac{1}{t} l n (\frac{P_{0}}{P_{0} - x}) = 0.02 m i n^{- 1} = 1.2 h r^{- 1} .$

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The correct option is B 1.2 A(g)→2B(g)+C(g)Let the initial pressure be P0 0 0 After 10 mins, (P0−x) 2x xAfter long time (t→∞) 0 2P0 P0Now, (P0−x)+2x+x+ vapour pressure of H2O=188.P0+2x=160 and 3P0+28=388So, P0=120 and x=20torr.k=1tln(P0P0−x)=0.02min−1=1.2hr−1.