The rate law for the reaction `2A to ` products is given by :
rate` =9.0 xx10^(-5)`A mol `L^(-1) S^(-1)`. What is the decompostion rate when `A=0.250` mol / litre ?

The rate equation for the reaction 2A + B rarr products is rate = k[A][B]^(2) . If k at T (K) is 5.0 xx 10^(-6) mol^(-2) L^(2)s^(-1) , the initial rate of the reaction, when [A] = 0.05 mol L^(-1) and [B] = 0.1 mol L^(-1) is

The rate constant for the reaction, 2N_2O_5 to 4NO_2 + O_2" is "4.0 xx 10^(-5)sec^(-1) . If the rate is 3.4 xx 10^(-5)mol.l^(-1)s^(-1) . Then the concentration of N_2O_5 (in mol litre^(-1) ) is :

The rate of a reaction is 1.2xx10^-3 L mol^-1 s^-1 . What is the order of the reaction?

For a first order for the reaction A rarr products : the rate of reaction at [A]=0.2 M is 1.0 xx 10^(-3) mol L^(-1)s^(-1) . The reaction will occur to 75% completion in .

The rate of a reaction is 1.2xx10^-3 L mol^-1s^-1 , What is the order of the reaction?

For a reaction a, A to Products , the units of rate constant are given as L mol^(-1) s^(-1) . Write the rate expression .

The rate law expression of a reaction 2A+B rarr A_(2)B is rate = k[A]^(2) with rate constant k =0.65 mol^(-1) L s^(-1) calculate rate of reaction when (a) [A]=0.25 mol L^(-1) [B]=1.84 mol L^(-1)

The rate law for the reaction A+B to Product, is given by the expression k [A] [B] . If the concentration of B is increased from 0.1 to 0.3 mol/L, keeping the value of A at 0.1 mol/L, the rate constant will be: