TG EAMCET 2 May 2025 Shift 1 Paper

Freundlich adsorption isotherm equation:$\frac{x}{m} = k p^{\frac{1}{n}}$where$\frac{x}{m} = \text{ mass of gas adsorbed per unit mass of adsorbent, }$$p =$ equilibrium pressure,$k$ and $n =$ empirical constants,$n > 1.$Or, taking logarithms,$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$Option A:$\frac{x}{m} = k p^{\frac{1}{n}} \;\; (n>1)$Correct.Option B:Extent of adsorption of gas is more at high temperature than at low temperature.Adsorption is an exothermic process.Hence, increasing temperature decreases adsorption (Le Chatelier's principle).Incorrect statement.Option C:$\frac{1}{n}$ represents the slope of the isotherm (in log-log plot)Since $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$,Correct.Option D:$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$ holds good over a limited range of pressures.At very high or very low pressures, Freundlich isotherm deviates.Correct.Final Answer:Option B - Extent of adsorption of gas is more at high temperature than at low temperature - is not correct about Freundlich adsorption isotherm.