Calculating Reaction Speed Changes: Second and First Order Kinetics at 298K

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The discussion focuses on calculating the reaction speed changes for the reaction A+B -> Products, which is second order with respect to A and first order with respect to B at a temperature of 298 K. Given initial concentrations of A at 0.75 mol L^-1 and B at 0.8 mol L^-1, the initial consumption speed is 0.00347 mol L^-1 min^-1. The impact of diluting the reaction mixture with additional water is analyzed, specifically how the concentrations of A and B change when 200 ml of water is added to 100 ml of the mixture and when 300 ml of water is added to 200 ml of the reaction. The equation for reaction speed is confirmed as V=k*[A]^2*[B].

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The reaction A+B ->Products is of second order related to A ,and of first order related to B.The reaction occurs on 298 K.When A=0.75 mol L^-1 and B=0,8 mol l^-1 the speed of the consumption of A and B is 0.00347 mol l^-1 min ^-1.What will be the speed of consumption of A and B after these changes happen: in 100 ml of the mixture we pour 200 ml of clean water?And in 200 ml of the reaction we add 300 ml of clean water?

So the equation of the speed must look like : V=k*[A]^2*...how do I relate this to the changes//?
 
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Elaia06 said:
The reaction A+B ->Products is of second order related to A ,and of first order related to B.The reaction occurs on 298 K.When A=0.75 mol L^-1 and B=0,8 mol l^-1 the speed of the consumption of A and B is 0.00347 mol l^-1 min ^-1.What will be the speed of consumption of A and B after these changes happen: in 100 ml of the mixture we pour 200 ml of clean water?And in 200 ml of the reaction we add 300 ml of clean water?

So the equation of the speed must look like : V=k*[A]^2*...how do I relate this to the changes//?


Your equation is OK. Use it. Frankly if there is a problem with this question it is that they are actually giving you more information than you need to solve it.
 

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