For Freundlich adsorption isotherm, a plot of log (x/m) (y-axis) and log p (x-axis) gives a straight line. The intercept and slope for the line is 0.4771 and 2, respectively. The mass of gas, adsorbed per gram of adsorbent if the initial pressure is 0.04 atm, is ______ × 10 –4 g.
(log 3 = 0.4771)

Solution

Chemistry ma’am · 4 steps

The intercept gives log k and the slope gives 1/n. Put both back into x/m = kp^(1/n).

1. Take logs. The Freundlich isotherm is x/m = kp^(1/n). Taking log on both sides gives a straight line in log p.

   log x/m = log k + 1/n log p

2. Read the line. The intercept is log k = 0.4771 = log 3, so k = 3. The slope is 1/n = 2.
3. Put in numbers. x/m = 3 × (0.04)² = 3 × 0.0016 = 0.0048 g per gram of adsorbent.
4. So. 0.0048 = 48 × 10⁻⁴, so the answer is 48.

The common slip: 6000. Taking the slope as n instead of 1/n gives 3 × (0.04)^(1/2) = 0.6, which is 6000, not 48.

The idea to keep. In a log(x/m) against log p plot, the slope is 1/n and the intercept is log k. Always take the antilog of the intercept before using it.

Answer from the NTA final answer key. Solution by PrepWiser.