For the reaction X → Y + Z, the rate constant is 0.00058 s-1. What percentage of X will be decomposed in 50 minutes?

Options

A. 90.02 %
B. 82.44 %
C. 88.82 %
D. 82.67 %

Answer

To solve the problem, we need to determine the percentage of reactant X that decomposes into products Y and Z after a given time, using the first-order rate constant provided.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Rate constant (k) = 0.00058 s⁻¹
- Time (t) = 50 minutes

2. **Convert Time to Seconds:**
- Since the rate constant is in seconds, we need to convert 50 minutes into seconds.
\[ t = 50 \text{ minutes} \times 60 \text{ seconds/minute} = 3000 \text{ seconds} \]

3. **Use the First-Order Reaction Formula:**
- For a first-order reaction, the relationship between the rate constant, time, and the concentration of reactants is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{100}{100 - a} \right) \]
where \( a \) is the percentage of X decomposed.

4. **Rearranging the Formula:**
- We can rearrange the formula to solve for \( a \):
\[ \log \left( \frac{100}{100 - a} \right) = k \cdot t \cdot \frac{2.303}{1} \]

5. **Substituting the Values:**
- Substitute \( k = 0.00058 \, \text{s}^{-1} \) and \( t = 3000 \, \text{s} \):
\[ \log \left( \frac{100}{100 - a} \right) = 0.00058 \times 3000 \]
\[ \log \left( \frac{100}{100 - a} \right) = 1.734 \]

6. **Calculating the Logarithm:**
- Now, we can calculate:
\[ \frac{100}{100 - a} = 10^{1.734} \]
- Calculating \( 10^{1.734} \):
\[ 10^{1.734} \approx 54.3 \]

7. **Setting Up the Equation:**
- Now we have:
\[ 100 = 54.3(100 - a) \]
- Expanding this gives:
\[ 100 = 5430 - 54.3a \]

8. **Solving for \( a \):**
- Rearranging the equation:
\[ 54.3a = 5430 - 100 \]
\[ 54.3a = 5330 \]
\[ a = \frac{5330}{54.3} \approx 98.1\% \]

9. **Calculating Percentage Decomposed:**
- Therefore, the percentage of X that has decomposed is approximately:
\[ a \approx 98.1\% \]

### Final Answer:
The percentage of X that will be decomposed in 50 minutes is approximately **98.1%**.