In a first order reaction, the concentrations of the reactant, 30 minutes and 40 minutes after the start are `C_(1)` and `C_(2) (` in moles `//` litre `)` respectively . What was `C_(0)`, initial concentration ?

Options

A. `C_(0)=[(C_(1)^(3))/(C_(2)^(4))]`
B. `C_(0)=[(C_(1))/(C_(2))]^(4)`
C. `C_(0)=[(C_(1))/(C_(2))]^(3)`
D. `C_(0)=[(C_(1)^(4))/(C_(2)^(3))]`

Explanation

To solve the problem, we need to determine the initial concentration \( C_0 \) of a first-order reaction given the concentrations at two different times.

### Step-by-Step Solution:

1. **Understand the First-Order Reaction Equation**:
For a first-order reaction, the concentration of the reactant at any time \( t \) can be expressed as:
\[ C_t = C_0 e^{-kt} \]
where \( C_t \) is the concentration at time \( t \), \( C_0 \) is the initial concentration, and \( k \) is the rate constant.

2. **Set Up the Equations**:
At \( t = 30 \) minutes, the concentration is \( C_1 \):
\[ C_1 = C_0 e^{-k \cdot 30} \]
At \( t = 40 \) minutes, the concentration is \( C_2 \):
\[ C_2 = C_0 e^{-k \cdot 40} \]

3. **Take the Natural Logarithm**:
From the equations above, we can express \( k \) in terms of \( C_0 \):
\[ \ln C_1 = \ln C_0 - k \cdot 30 \]
\[ \ln C_2 = \ln C_0 - k \cdot 40 \]

4. **Rearranging the Equations**:
Rearranging these gives us:
\[ k \cdot 30 = \ln C_0 - \ln C_1 \quad \text{(1)} \]
\[ k \cdot 40 = \ln C_0 - \ln C_2 \quad \text{(2)} \]

5. **Divide the Two Equations**:
Dividing equation (1) by equation (2):
\[ \frac{k \cdot 30}{k \cdot 40} = \frac{\ln C_0 - \ln C_1}{\ln C_0 - \ln C_2} \]
Simplifying gives:
\[ \frac{3}{4} = \frac{\ln C_0 - \ln C_1}{\ln C_0 - \ln C_2} \]

6. **Cross-Multiply**:
Cross-multiplying gives:
\[ 3(\ln C_0 - \ln C_2) = 4(\ln C_0 - \ln C_1) \]
Expanding this leads to:
\[ 3 \ln C_0 - 3 \ln C_2 = 4 \ln C_0 - 4 \ln C_1 \]

7. **Rearranging Terms**:
Rearranging gives:
\[ 4 \ln C_1 - 3 \ln C_2 = \ln C_0 \]

8. **Exponentiate to Solve for \( C_0 \)**:
Exponentiating both sides results in:
\[ C_0 = \frac{C_1^4}{C_2^3} \]

### Final Answer:
Thus, the initial concentration \( C_0 \) is given by:
\[ C_0 = \frac{C_1^4}{C_2^3} \]