The following results were obtained during kinetic studies of the reaction. 2A + B → Products [MAINS Esperment [B] [A] (in mol L) (in mol L) Initial Rate of reaction (in mol Lmin (D 0.10 0.20 6.93 x 10- (I) 0.10 0.25 6.93 x 10 (IT) 0.20 0.30 1.386 x 10-2 The time (in minutes) required to consume half of A is :

Compound 'A' and B react according to the following chemical equation. A_((g)) + 2B_((g)) to 2C_((g)) Concentration of either A or B were changed keeping the concentrations of one of the reactants constant and rates were mesured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction. {:("Experiment","Initial concentration of"[A]//mol L^(-1),"Initial concentration of"[B]mol L^(-1),"Initial rate of formation of"[c]//mol L^(-1)s^(-1)),(1,0.30,0.30,0.10),(2,0.30,0.60,0.40),(3,0.60,0.30,0.20):}

Compound 'A' and B react according to the following chemical equation. A_((g)) + 2B_((g)) to 2C_((g)) Concentration of either A or B were changed keeping the concentrations of one of the reactants constant and rates were mesured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction. {:("Experiment","Initial concentration of"[A]//mol L^(-1),"Initial concentration of"[B]mol L^(-1),"Initial rate of formation of"[c]//mol L^(-1)s^(-1)),(1,0.30,0.30,0.10),(2,0.30,0.60,0.40),(3,0.60,0.30,0.20):}

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

For a hypothetical reaction A + B rarr C , suggest the rate law and order form the following data: |{:("Experiment",[A] (mol L^(-1)),[B] (mol L^(-1)),"Refer of reaction" (mol L^(-1) s^(-1)),),(I,0.25,0.25,3.0 xx 10^(-3),),(II,0.50,0.25,6.0 xx 10^(-3),),(III,0.50,0.50,1.20 xx 10^(-2),):}|

For a hypothetical reaction A + B rarr C , suggest the rate law and order form the following data: |{:("Experiment",[A] (mol L^(-1)),[B] (mol L^(-1)),"Refer of reaction" (mol L^(-1) s^(-1)),),(I,0.25,0.25,3.0 xx 10^(-3),),(II,0.50,0.25,6.0 xx 10^(-3),),(III,0.50,0.50,1.20 xx 10^(-2),):}|

The following data are obtained for a reaction, X+Yrarr"Products" . {:("Expt. ",[X_(0)]//"mol ",[Y_(0)]//"mol ","rate/mol L"^(-1)s^(-1)),(1,0.25,0.25,1.0xx10^(-6)),(2,0.50,0.25,4.0xx10^(-6)),(3,0.25,0.50,8.0xx10^(-6)):} The overall of the reaction is

In a reaction A rarr Products , the concentration of A decreases from 0.6 mol L^(-1) to 0.3 mol L^(-1) in 10 minutes. Calculate the rate during this time interval.

In a reaction 2A + B to 3C , the concentration of A decreases from 0.5 mol L^(-11) to 0.3 mol L^(-1) in 10 minutes. The rate of production of 'C during this period is: