A plot of log x/m versus log p for the a...

A plot of log x/m versus log p for the adsorption of a gas on a solid gives a straight line with slope equal to

1/n

log K

`-log K`

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A plot of log (x/m) against log P for the adsorption of a gas on a sloid gives a straight a straight line with slope equal to :

For adsorption of a gas on a solid, the plot of log x/m vs log P is linear with a slope equal to (n being a whole number)

For a linear plot of log (x/m) versus log P in a Freundlich adsorption isotherm, which of the following statement is correct? (k and n are constants).

Rate constant 'k' of a reaction varies with temperature 'T' according to the equation: log k= log A- (E_a)/(2.303R) (1/T) where E_a is the activation energy. When a graph is plotted for log k vs. 1/T , a straight line with a slope of - 4250 K is obtained. Calculate E_a for the reaction (R = 8.314 JK^-1 mol^-1)

Rate constant, k of a reaction varies with temperature as: log k = Constant- E_a/(2.303RT) where E_a is the activation energy. When a Graph is plotted for log k vs 1/T ,a straight line with a slope of -6670K is obtained. Calculate the energy of activation for this reaction. ( R = 8.314 JK^-1 mol^-1 )

Read the given passage and answers following questions : There are mainly two type of adsorption of gases on solids. If accumulation of gases on the surface of solid occurs on account of weak vander waal forces, the adsorption is termed as physical adsorption. When gas molecules or atoms held to solid surface by chemical bonds, adsorption is termed as chemical adsorption. The chemical bonds may be covalent or ionic. Chemical adsorption involves a high energy of activation therefore it is referred as Activated adsorption. A physical adsorption at low temperature may pass into chemical adsorption temperature is increased. Give conditions of temperature which favours physical adsorption.

For Freundlich isotherm, a graph of log x/m is plotted against log P. The slope of the line and its y-axis intercept respectively correspond to

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