Half life (`t_1`) of the first order reaction and half life (`t_2`) of the 2`""^(nd)` order reaction are equal. If initial concentration of reactants in both reaction are same then ratio of the initial rate of the reaction

Options

A. 1
B. 2
C. 0.693
D. 1.44

Answer

To solve the problem, we need to find the ratio of the initial rates of a first-order reaction and a second-order reaction given that their half-lives are equal and the initial concentrations of the reactants are the same.

### Step-by-Step Solution:

1. **Understanding Half-Life of Reactions:**
- The half-life of a first-order reaction (\(t_1\)) is given by the formula:
\[ t_1 = \frac{0.693}{k_1} \]
- The half-life of a second-order reaction (\(t_2\)) is given by:
\[ t_2 = \frac{1}{k_2[A]_0} \]
where \([A]_0\) is the initial concentration of the reactant.

2. **Setting the Half-Lives Equal:**
- Since it is given that \(t_1 = t_2\), we can equate the two expressions:
\[ \frac{0.693}{k_1} = \frac{1}{k_2[A]_0} \]

3. **Rearranging the Equation:**
- Rearranging gives us:
\[ k_1 = 0.693 \cdot k_2 \cdot [A]_0 \]

4. **Writing the Rate Equations:**
- The initial rate of the first-order reaction (\(R_1\)) is:
\[ R_1 = k_1[A]_0 \]
- The initial rate of the second-order reaction (\(R_2\)) is:
\[ R_2 = k_2[A]_0^2 \]

5. **Finding the Ratio of Initial Rates:**
- The ratio of the initial rates \(R_1/R_2\) can be expressed as:
\[ \frac{R_1}{R_2} = \frac{k_1[A]_0}{k_2[A]_0^2} \]
- Simplifying this gives:
\[ \frac{R_1}{R_2} = \frac{k_1}{k_2[A]_0} \]

6. **Substituting for \(k_1\):**
- From our earlier step, we substitute \(k_1\):
\[ \frac{R_1}{R_2} = \frac{0.693 \cdot k_2 \cdot [A]_0}{k_2[A]_0} \]
- The \(k_2\) and \([A]_0\) cancel out:
\[ \frac{R_1}{R_2} = 0.693 \]

### Final Answer:
The ratio of the initial rates of the reactions is:
\[ \frac{R_1}{R_2} = 0.693 \]