# Why does mixing two solutions decrease the concentration of the solutes?

The irreversible liquid reaction

$$\ce{A + B -> C + D}$$

follows an elementary rate law and is carried out isothermally in a plug flow reactor (PFR). The volume of the PFR is 500 litres. The concentration of $$\ce{A}$$ is $$\pu{2 mol L-1}$$ and concentration of $$\ce{B}$$ is $$\pu{2 mol L-1}$$ before mixing. The volumetric flow rate of each stream is $$\pu{5 L min-1}.$$

The streams are mixed immediately before entering. The reaction rate constant at $$\pu{20 °C}$$ is $$\pu{0.05 L mol-1 min-1}$$ and $$E = \pu{20 kcal mol-1}.$$ (Note: gas constant $$R = \pu{1.987 cal mol-1 K-1})$$

Find conversion at $$\pu{40 °C}.$$ From the solution given the $$c_{\ce{A},0}$$ used is $$\pu{1 mol L-1}$$ which is after mixing. Need some clarification on this. Why is it not $$\pu{2 mol L-1}$$ before mixing?

• From the solution given the Cao used is 1mol/L which is after mixing. Need some clarification on this. Why is it not 2mol/L before mixing – user80057 Jun 12 '19 at 9:18
• Because it has been mixed. Each substance is now in doubled volume, compared to volume before mixing. If you had whiskey and if you mixed it with the same volume of soda, it wouldn't have the same strength as before, would it ? – Poutnik Jun 12 '19 at 16:25
• alright.. thanks for the explanation. – user80057 Jun 18 '19 at 9:45