A Freundlich adsorption isotherm plot of log(x/m) versus log(P) is a straight line with slope 1/2. At a pressure of 0.5 atm, the Freundlich constant k = 10. What is the amount of solute adsorbed per gram of adsorbent?

Question

Accepted Answer

1 g. Freundlich isotherm: x/m = k * P^(1/n). Here 1/n = 1/2 (slope), k = 10, P = 0.5 atm. x/m = 10 * (0.5)^(1/2) = 10 * sqrt(0.5) = 10/sqrt(2) ≈ 7.07. But the answer 7 is approximately 7. Alternatively if k is the intercept of log(x/m) vs log(P): log(x/m) = (1/2)*log(P) + log(k). At P = 0.5: log(x/m) = (1/2)*log(0.5) + log(10) = (1/2)*(-0.301) + 1 = -0.15 + 1 = 0.85. x/m = 10⁰.85 ≈ 7.08 ≈ 7. So the answer is approximately 7 g/g.

A Freundlich adsorption isotherm plot of log(x/m) versus log(P) is a straight line with slope 1/2. At a pressure of 0.5 atm, the Freundlich constant k = 10. What is the amount of solute adsorbed per gram of adsorbent?

Solution

Related JEE Advanced Chemistry questions