Surface ChemistryhardNUMERICAL

For the Freundlich adsorption isotherm, a plot of against gives a straight line making an angle of wSurface Chemistry Chemistry Question

Question

For the Freundlich adsorption isotherm, a plot of $\log(x/m)$ against $\log p$ gives a straight line making an angle of $45^\circ$ with the positive x-axis. If the intercept on the y-axis is $0.6020$, calculate the amount of gas adsorbed per gram of the adsorbent ($x/m$ in grams) at a pressure of $2.5\text{ atm}$. (Take $\log 2 = 0.3010$)

Answer: 10

💡 Solution & Explanation

The logarithmic form of the Freundlich equation is $\log(x/m) = \log k + \frac{1}{n}\log p$. The slope of the plot is $\frac{1}{n} = \tan(45^\circ) = 1$. The intercept is $\log k = 0.6020$. Since $\log 2 = 0.3010$, $\log 4 = 2 \times 0.3010 = 0.6020$, which implies $k = 4$. Using the isotherm equation $\frac{x}{m} = k p^{1/n}$, substitute the values: $\frac{x}{m} = 4 \times (2.5)^1 = 10\text{ g}$.

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Advanced Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry