Rate constant for first order reaction is `5.78xx10^(-5)S^(-1)`. What percentage of initial reactant will react in 10 hours ?

Options

A. `12. 5%`
B. `25%`
C. `87. 5%`
D. `75%`

Explanation

To solve the problem of determining what percentage of the initial reactant will react in 10 hours for a first-order reaction with a given rate constant, we can follow these steps:

### Step 1: Identify the given values
- Rate constant (K) = \(5.78 \times 10^{-5} \, \text{s}^{-1}\)
- Time (T) = 10 hours

### Step 2: Convert time from hours to seconds
Since the rate constant is in seconds, we need to convert the time from hours to seconds:
\[ T = 10 \, \text{hours} \times 3600 \, \text{s/hour} = 36000 \, \text{s} \]

### Step 3: Use the first-order reaction formula
The formula for a first-order reaction is given by:
\[ K = \frac{2.303}{T} \log \left( \frac{A_0}{A} \right) \]
Where:
- \(A_0\) = initial concentration
- \(A\) = concentration at time \(T\)

### Step 4: Rearrange the formula to solve for \(\log \left( \frac{A_0}{A} \right)\)
Rearranging the formula gives:
\[ \log \left( \frac{A_0}{A} \right) = K \cdot T \cdot \frac{2.303}{1} \]

### Step 5: Substitute the values into the equation
Substituting the values of K and T:
\[ \log \left( \frac{A_0}{A} \right) = 5.78 \times 10^{-5} \, \text{s}^{-1} \times 36000 \, \text{s} \times 2.303 \]

### Step 6: Calculate the value
Calculating the right-hand side:
\[ \log \left( \frac{A_0}{A} \right) = 5.78 \times 10^{-5} \times 36000 \times 2.303 \approx 0.968 \]

### Step 7: Take the antilog to find \(\frac{A_0}{A}\)
Taking the antilog:
\[ \frac{A_0}{A} = 10^{0.968} \approx 8.06 \]

### Step 8: Solve for A
Rearranging gives:
\[ A = \frac{A_0}{8.06} \]

### Step 9: Calculate the percentage of reactant that has reacted
To find the percentage of the initial reactant that has reacted:
\[ \text{Percentage reacted} = \left(1 - \frac{A}{A_0}\right) \times 100 = \left(1 - \frac{1}{8.06}\right) \times 100 \]
Calculating this gives:
\[ \text{Percentage reacted} = \left(1 - 0.124\right) \times 100 \approx 87.5\% \]

### Final Answer
The percentage of the initial reactant that will react in 10 hours is approximately **87.5%**.

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