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Standard XII

Chemistry

Fruendlich Isotherm

20% surface s...

Question

$20 %$ surface sites have absorbed $N_{2}$ . On heating, $N_{2}$ gas is evolved from sites and were collected at $0.001$ atm and $298$ K in a container of volume $2.46 c m^{3}$ . Density of surface sites is $6.023 \times 10^{14}$ $c m^{- 2}$ and surface area of $1000 c m^{2}$ . Find out the number of surface sites occupied per molecule of $N_{2}$ .

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Solution

Given that for absorbed $N_{2}$ on surface sites,

$P_{N_{2}} = 0.001 a t m, V = 2.46 c m^{3} = 2.46 \times 10^{- 3} l, T = 298 K$

$∴ n_{N_{2}} = \frac{P V}{R T} = \frac{0.001 \times 2.46 \times 10^{- 6}}{0.0821 \times 298} = 1.0 \times 10^{- 7}$

Molecules of absorbed $N_{2} = 1.0 \times 10^{- 7} \times 6.023 \times 10^{23}$ $= 6.023 \times 10^{16}$

Total surface sites available $=$ Number of sites per $c m^{2} \times$ Area $= 60.23 \times 10^{14} \times 1000 = 6.023 \times 10^{17}$

Surface sites on which $N_{2}$ is absorbed $=$ $20 %$ $\times$ Available sites

$= \frac{20}{100} \times 6.023 \times 10^{17}$ $= 12.046 \times 10^{16}$
$∴$ Number of sites absorbed per molecule of $N_{2} = \frac{12.046 \times 10^{16}}{6.023 \times 10^{16}} = 2$

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20% surface sites have absorbed N2. On heating, N2 gas is evolved from sites and were collected at 0.001 atm and 298 K in a container of volume 2.46cm3. Density of surface sites is 6.023×1014 cm−2 and surface area of 1000cm2. Find out the number of surface sites occupied per molecule of N2.