Error in calculation of rate of reaction through graphs
Am I missing something?


Yes. You are missing the rate law and everything that goes along with it. It is only easy to fit zeroth order raw rate data. To produce linear fits, you need to consider the possibility that your reaction has a first or second order rate.

In addition to rate being defined as the (microscopic) change in concentration respect to time, you can also derive a rate law representing this change since the rate of the reaction is going to be proportional to things like the number of collisions between particles (which increases as there are more particles, i.e. higher concentration). So, we get a rate law that is something like the following:

$$\mathrm{rate}=kC^x$$

where $k$ is the rate constant, which is dependent on things like temperature and the activation barrier for the reaction, and $x$ defines the order of reaction, which is dependent on the mechanism and can be loosely interpreted as being related to the number of particles the must collide in the fundamental step(s). The value of $x$ is determined empirically (one such way is the experiment you did), though common values for simple reactions are integers, with $x=0,1,2$ being most common.

With both equations for rate in hand, we can do some to obtain integrated rate laws, which take the form $C(t)$: just the sort of equations that you are looking for. Note the minus signs in my equations. Rate is usually expressed based on consumption of reactants. 


  $x=0$, zeroth order


$$\begin{align}-\dfrac{dC}{dt}&amp;=kC^0 = k\\
-dC &amp;= kdt \\
-\int_{C_0}^{C} dC &amp;= \int_0^t kdt \\
C - C_0 &amp;= -kt \\
C &amp;= -kt + C_0
\end{align}$$

A reaction with a zeroth order rate will produce a graph of $C(t)$ that is linear.


  $x=1$, first order


$$\begin{align}-\dfrac{dC}{dt}&amp;=kC^1 = kC\\
-\dfrac{dC}{C} &amp;= kdt \\
-\int_{C_0}^{C} \dfrac{1}{C} dC &amp;= \int_0^t kdt \\
\ln C - \ln C_0 &amp;= -kt\\
\ln C &amp;= -kt + ln C_0 \\
C &amp;= C_0 e^{-kt}
\end{align}$$

A reaction with a first order rate will produce a graph of $C(t)$ that shows exponential decay. However, my derivation provides a linear equation: $\ln C = -kt + \ln C_0$. If you determine the natural log of each concentration and plot against $t$, you will get a line (if the reaction is first order).


  $x=2$, second order


$$\begin{align}-\dfrac{dC}{dt}&amp;=kC^2\\
-\dfrac{dC}{C^2} &amp;= kdt \\
-\int_{C_0}^{C} \dfrac{1}{C^2}dC &amp;= \int_0^t kdt \\
\dfrac{1}{C} - \dfrac{1}{C_0} &amp;= -kt \\
\dfrac{1}{C} &amp;= -kt + \dfrac{1}{C_0} \\
C &amp;= -\dfrac{C_0}{1-C_0 kt}
\end{align}$$

A reaction with a second order rate will produce a graph of $C(t)$ that shows power decay. However, my derivation provides a linear equation: $\dfrac{1}{C} = -kt + \dfrac{1}{C_0}$. If you determine the inverse of each concentration and plot against $t$, you will get a line (if the reaction is second order).