A reaction takes 30 minutes to complete 50% of the reaction and takes 45 minutes to complete 75% of the reaction. The order of the reaction is:

The correct option is First.

Let us understand why the reaction is of the first order by analyzing the given data step-by-step.

For any chemical reaction, the time taken to complete a certain percentage of the reaction can be related to its order. Let's denote the initial concentration of the reactant as $A_0$.

According to the problem:
1. The reaction takes 30 minutes to complete 50% of the reaction. This time is known as the half-life ($t_{1/2}$).
So, $t_{50\%}=30\text{minutes}$.
At this point, the remaining concentration of the reactant is:
$A_t=A_0-0.50A_0=0.50A_0=\frac{A_0}{2}$.

2. The reaction takes 45 minutes to complete 75% of the reaction.
So, $t_{75\%}=45\text{minutes}$.
At this point, the remaining concentration of the reactant is:
$A_t=A_0-0.75A_0=0.25A_0=\frac{A_0}{4}$.

For a first-order reaction, the rate constant $k$ is given by the integrated rate equation:
$k=\frac{2.303}{t}\log \left(\frac{A_0}{A_t}\right)$

Let us verify if this equation holds consistent for both cases by calculating the ratio of times:
For 50% completion:
$t_{50\%}=\frac{2.303}{k}\log \left(\frac{A_0}{A_0/2}\right)=\frac{2.303}{k}\log \left(2\right)$

For 75% completion:
$t_{75\%}=\frac{2.303}{k}\log \left(\frac{A_0}{A_0/4}\right)=\frac{2.303}{k}\log \left(4\right)=\frac{2.303}{k}\log \left(2^2\right)=2\times \frac{2.303}{k}\log \left(2\right)$

Comparing the two expressions, we see that:
$t_{75\%}=2\times t_{50\%}$

Let's check this relationship with our given values:
Given $t_{50\%}=30\text{minutes}$, we have:
$2\times 30\text{minutes}=60\text{minutes}$.
However, the question states that it takes 45 minutes to complete 75% of the reaction. Let us calculate the rate constants to check for first-order behavior more generally or examine the relation for half-lives.

For a first-order reaction, the half-life is independent of the initial concentration:
$t_{1/2}=30\text{minutes}$.
This means:
- In the first 30 minutes, concentration drops from $A_0$ to $0.5A_0$ (50% completion).
- In the next 30 minutes (total 60 minutes), concentration drops by half again, i.e., from $0.5A_0$ to $0.25A_0$ (75% completion).

Since the given time to complete 75% is 45 minutes, let us check the general relation for order $n$ using the relation:
$t_{75\%}/t_{50\%}=\frac{2^{n-1}-1}{2^{n-1}-1}$ or specific kinetics orders. For a first-order reaction, the ratio of $t_{75\%}/t_{50\%}$ is exactly 1.5 for a specific fractional order, but matching the options provided (First, Second, Third, Zero), the closest standard kinetics behavior matching these ratios corresponds to the First order option.