Consider the data [X] [Y] [Z] initial rate M M M M · s −1 Exp 1 0.30 0.20 0.35 0.210 Exp 2 0.60 0.10 0.70 0.420 Exp 3 0.60 0.20 0.70 0.420 Exp 4 0.60 0.40 0.35 0.105 What is a correct rate law for the reaction?

Rate = k [X]⁻¹ [Z]²

Explanation:

[X] [Y] [Z] initial rate M M M M · s −1

Exp 1 0.30 0.20 0.35      0.210

Exp 2 0.60 0.10 0.70      0.420

Exp 3 0.60 0.20 0.70     0.420

Exp 4 0.60 0.40 0.35      0.105

In Experiment 2 and 3 where the concentrations of Y and Z were constant, doubling the concentration of Y had no effect on the rate of the reaction. This means, that the rate of the reaction is zero order with respect to Y.

In experiment 3 and 4, dividing the concentration of Z by 2, causes the rate of the reaction to decrease by 4. This means the rate of the reaction is second order with respect to Z.

In experiment 1 and 4, doubling the concentration of X, causes the rate of the reaction to decrease by half. This means that X has an order of -1 with respect to the rate of the reaction.

The rate expression is given as;

Rate = k [X]⁻¹[Y]⁰[Z]²

Rate = k [X]⁻¹ [Z]²

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.