10% sites of catalyst bed have absorbed by H2. On heating H2 gas is evolved from sites and collected at 0.03 atm and 300 K in a small vessel of 2.46 cm3. Number of sites available is 5.4×1017 per cm2 and surface area is 100cm2. Find out the no. of surface sites occupied per molecule of H2. (Given NA=6×1023)

V=2.46CC=2.461000LP=0.03atm T=300Kn=PVRT=0.03×2.460.0821×300×1000=3×10−6NH2=3×10−6×6.023×1023=18×1017=1.8×1018 Number of sites ( active )=5.4×1017×10×100100=5.4×1018 Number of sites per H2 molecule =5.4×10181.8×1018=3

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10% sites of catalyst bed have absorbed by H₂. On heating H₂ gas is evolved from sites and collected at 0.03 atm and 300 K in a small vessel of 2.46 cm³. Number of sites available is 5.4 $\times$ 10¹⁷ per cm² and surface area is 100cm². Find out the no. of surface sites occupied per molecule of H₂. (Given $N_{A} = 6 \times 10^{23}$ )

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Detailed Solution

$\begin{array}{l} V = 2.46 CC = (\frac{2.46}{1000}) LP = 0.03 atm T = 300 K \\ n = \frac{PV}{RT} = \frac{0.03 \times 2.46}{0.0821 \times 300 \times 1000} = 3 \times 10^{- 6} \\ N_{H_{2}} = 3 \times 10^{- 6} \times 6.023 \times 10^{23} = 18 \times 10^{17} = 1.8 \times 10^{18} \\ Number of sites (active) = \frac{5.4 \times 10^{17} \times 10 \times 100}{100} = 5.4 \times 10^{18} \\ Number of sites per H_{2} molecule = \frac{5.4 \times 10^{18}}{1.8 \times 10^{18}} = 3 \end{array}$