**Question 1:**

The concentration of the reactant in a first-order chemical reaction decreases with time as it is consumed to form products and the concentration of [A] varies with time as follows

**[A] = [A] _{0} e^{-kr}**

Where [A]_{0} is the initial concentration of the reactant at time t = 0, [A] is the concentration of the reactant at time t, and k is the rate constant for this particular reaction. Sketch the shape of the graph of [A]/[A]_{0} plotted as the y-axis against time t plotted as the x-axis and label the two axes. You do not need to calculate anything. Below Figure shows the shape of a negative exponential curve.

**Question 2:**

**[A] = [A] _{0} e^{-kt}**

Write the natural-log form for a kinetic first-order reaction with ([A]/[A]_{0}) as the subject.

**Solution 1: **

**[A]= [A] _{0} e^{-kt} or, [A]/ [A]_{0} = e^{-kt}**

Remember that k is a constant for a given reaction at a fixed temperature and plotting [A]/[A]_{0} versus t is a negative exponential and has the shape sketched in below Figure which has an intercept at [A]/[A]_{0} = 1 that is [A] = [A]_{0} at t = 0. Note that using this non-linear plot it is not easy to deduce the rate constant k.

**Solution 2:**

**[A]= [A] _{0} e^{-kt} **

From the first order kinetics equation: (1) take natural logs throughout; (2) expand out the right hand side which is the log of a product of two terms into the sum of the log of each term;

**ln [A]= ln {[A] _{0} e^{-kt} }**

**ln [A]= {ln [A] _{0}} {ln e^{-kt} }**

(3) the natural log of the exponential is just the exponent -kt (4) move ln([A]_{0}) to the left hand side by subtracting ln([A]_{0}) from both sides; (5) convert the difference of two logs into the log of the ratio.

**ln [A] = ln [A] _{0} – kt **

**ln [A] – ln [A] _{0} = – kt**

**ln {[A]/[A] _{0}} = – kt**

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