For the empirical Freundlich adsorption isotherm, a plot of against yields a straight line inclined — Surface Chemistry Chemistry Question
Question
For the empirical Freundlich adsorption isotherm, a plot of $\log(x/m)$ against $\log P$ yields a straight line inclined at an angle of $45^\circ$. If the y-intercept is $0.6990$, what is the precise amount of gas adsorbed per gram of adsorbent at a pressure of $0.5\text{ atm}$?
Answer: C
💡 Solution & Explanation
Slope $= 1/n = \tan(45^\circ) = 1$. Intercept $= \log k = 0.6990 \implies k = 5$ (since $\log 5 = 0.6990$). Using the equation $x/m = kP^{1/n}$, we get $x/m = 5 \times (0.5)^1 = 2.5\text{ g}$.
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