For the following graphs,

Choose from the options given below, the correct one regarding order of reaction is:

For a zero-order reaction:

$$\mathrm{Rate = k[A]^0 = k}$$

the rate is independent of concentration.

Hence, in graph (a), rate remains constant with time, giving a horizontal line.

For zero-order reactions, half-life is:

$$\mathrm{t_{1/2} = \frac{[A]_0}{2k}}$$

Thus:

$$\mathrm{t_{1/2} \propto [A]_0}$$

Therefore, graph (b) gives a straight line passing through the origin and represents zero-order kinetics.

For a first-order reaction:

$$\mathrm{Rate = k[A]}$$

The integrated rate equation is:

$$\mathrm{[A] = [A]_0 e^{-kt}}$$

Thus, concentration decreases exponentially with time.

Hence, graph (c) represents first-order kinetics.

For first-order reactions:

$$\mathrm{Rate \propto [A]}$$

Therefore, graph (e), which is a straight line between rate and concentration passing through the origin, also represents first-order kinetics.

Graph (d) shows constant concentration with time.

This means concentration does not decrease and effectively no reaction occurs.

Therefore:

$$\mathrm{(a)\ and\ (b)\ \rightarrow\ Zero\ Order}$$

$$\mathrm{(c)\ and\ (e)\ \rightarrow\ First\ Order}$$