In the fig AOB is a straight line. Find the value of x
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)
In the fig. what value of x will make XOY a straight line
int e^x (cot x + log sin x)dx is equal to :
int e^x (tan x + log sec x)dx is equal to :
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)
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
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 )
