If 50% of a reaction occurs in 100 second and 75% of the reaction occurs in 200 second, the order of this reaction is:

We are given that 50% of the reaction occurs in 100 seconds, and 75% of the reaction occurs in 200 seconds.

Let the initial concentration be $$a$$. Then:

After 100 s: concentration remaining = $$\frac{a}{2}$$

After 200 s: concentration remaining = $$a - 0.75a = \frac{a}{4}$$

Notice that during the first 100 seconds, the concentration drops from $$a$$ to $$\frac{a}{2}$$ (i.e., it halves). During the next 100 seconds (from 100 s to 200 s), the concentration drops from $$\frac{a}{2}$$ to $$\frac{a}{4}$$ (it halves again).

So the half-life is the same in both intervals: $$t_{1/2} = 100$$ s in each case.

Now recall the key property of a first-order reaction: the half-life of a first-order reaction is independent of the initial concentration. The formula is:

$$t_{1/2} = \frac{0.693}{k}$$

Since the half-life remains constant at 100 s regardless of the concentration (it was 100 s when starting from $$a$$, and again 100 s when starting from $$\frac{a}{2}$$), this confirms that the reaction is first order.

Verification with the integrated rate law:

For a first-order reaction: $$\ln\frac{[A]_0}{[A]_t} = kt$$

At $$t = 100$$ s: $$\ln\frac{a}{a/2} = k \times 100$$

$$\ln 2 = 100k$$

$$k = \frac{\ln 2}{100} = \frac{0.693}{100} = 6.93 \times 10^{-3}$$ s$$^{-1}$$

At $$t = 200$$ s: $$\ln\frac{a}{a/4} = k \times 200$$

$$\ln 4 = 200k$$

$$k = \frac{\ln 4}{200} = \frac{2 \ln 2}{200} = \frac{\ln 2}{100} = 6.93 \times 10^{-3}$$ s$$^{-1}$$

The rate constant $$k$$ is the same in both cases, confirming first-order kinetics.

Why not zero order? For a zero-order reaction, $$t_{1/2} = \frac{a}{2k}$$, which depends on the initial concentration. If we start with concentration $$\frac{a}{2}$$, the new half-life would be $$\frac{a}{4k} = \frac{t_{1/2}}{2} = 50$$ s, not 100 s. So the half-life would change, which contradicts the data.

Why not second order? For a second-order reaction, $$t_{1/2} = \frac{1}{ka}$$. Starting from $$\frac{a}{2}$$, the half-life would become $$\frac{1}{k \cdot a/2} = \frac{2}{ka} = 2 \times 100 = 200$$ s, not 100 s. Again, the half-life changes, which contradicts the data.

Therefore, the reaction is first order.

The correct answer is Option D.