At 70 K, the adsorption of (g) on iron surface obeys Freundlich adsorption isotherm. The following data is collected experimentally: P = 4, 25, 64 bar and x/m = 0.2, 0.5, 0.8. The moles of (g) adsorbe

Question

At 70 K, the adsorption of $N_2$(g) on iron surface obeys Freundlich adsorption isotherm. The following data is collected experimentally: P = 4, 25, 64 bar and x/m = 0.2, 0.5, 0.8. The moles of $N_2$(g) adsorbed per g of iron at 36 bar and 70 K is

Prashant Kotian · Accepted Answer

Correct answer: B. Let's find k and n using x/m = k * P^(1/n). Substitute two points: log(0.2) = log k + (1/n)log 4; log(0.5) = log k + (1/n)log 25. Subtracting: log(2.5) = (1/n)log(6.25) => 1/n = 0.5 => n = 2. Then, 0.2 = k * 4^0.5 = 2k => k = 0.1. Thus, x/m = 0.1 * P^0.5. At P = 36 bar, x/m = 0.1 * 36^0.5 = 0.6 g. Moles of = 0.6 / 28 = 3/140.

💡 Solution & Explanation

Practice 22,000+ questions like this