The conversion of molecules X to Y follows second order kinetics. If concentration of x is increased to three times how will it affect the rate of formation of Y?
Because the reaction X Y has second-order kinetics, the rate law equation will be Rate = kC2Rate = kC2 , with C = [x]C = [x] . The rate law equation for the reaction X Y will be Rate = k C2Rate = k C2 , with C = [x]C = [x] because it possesses second-order kinetics. So, [x]=3C mol L - 1[x]=3C mol L - 1 The rate equation is Rate=K(3C)2Rate=K(3C)2 = 9(kC2)9(kC2) As a result, the reaction rate will increase by 9 times. As a result, the rate at which Y is formed will grow by 9 times.
