Rate constant `k` varies with temperature by equation `log k (min^(-1)) = log 5 - (2000 kcal)/(RT xx 2.303)`. We can conclude that

Rate constant k varies with temperature by equation, log k (min+^(-1))=5-(2000)/(T(K)) We can conclud:

Rate constant k varies with temperature by equation, log k (min+^(-1))=5-(2000)/(T(K)) We can conclud:

Rate constant K varies with temperature by equation log K("min"^(-1))=5-((2000))/T .We can conclude that (R=8.314J"mol"^(-1)K^(-1) (or) cal "mol"^(-1)K^(-1) )

Rate constant k of a reaction varies with temperature according to equation: log k= constant - (E_a)/( 2.303 R).(1)/(T ) What is the activation energy for the reaction. When a graph is plotted for log k versus 1/T a straight line with a slope-6670 K is obtained. Calculate energy of activation for this reaction (R=8.314 JK^(-1) mol^(-) 1)

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)

The rate constant 'k'. For a reaction varies with temperature 'T' according to the question. 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. Calcualte E_(a) for this reaction.

Rate constant k of a reaction varies with temperature according to the equation logk = constant -E_a/(2.303R)(1/T) where E_a is the energy of activation for the reaction . When a graph is plotted for log k versus 1/T , a straight line with a slope - 6670 K is obtained. Calculate the energy of activation for this reaction