The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Consider a certain reaction \(A\) \(→\) \(Products\) with \(k = 2.0 \times 10^{-2 }s^{-1}\) . Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L-1.
The decomposition of A into product has value of k as \(4.5 \times 10^3 s^{-1}\) at \(10°C\) and energy of activation \(60\ kJ mol^{-1 }\). At what temperature would k be \(1.5 \times 10^4 s^{-1}\)?
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
The rate constant for the first order decomposition of \(H_2O _2\) is given by the following equation:\(log\ k = 14.34 - 1.25 \times 10^4\ K/T \)Calculate \(E_a\) for this reaction and at what temperature will its half-period be 256 minutes?
The rate of a reaction quadruples when the temperature changes from \(293 \ K\) to \(313\ K\). Calculate the energy of activation of the reaction assuming that it does not change with temperature.
The decomposition of hydrocarbon follows the equation\(k = (4.5 \times 10^{11} s^{-1})e^{-28000 \ K/T }\)\(Calculate\ E_a\).
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with \(t_{\frac 12} = 3.00\ hours\). What fraction of sample of sucrose remains after \(8 \ hours\)?
The rate constant for the decomposition of \(N_2O_5\) at various temperatures is given below:
Draw a graph between ln k and \(\frac 1T\) and calculate the values of \(A\) and \(E_a\).Predict the rate constant at 30 ºC and 50 ºC.