The decomposition of `A` into product has value of `k` as `4.5xx10^(3)s^(-1)` at `10^(@)C` and energy of activation of `60kJmol^(-1)`. At what temperature would `k` be `1.5xx10^(4)s^(-1)?`

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and the activation energy (Ea). The equation is given by:

\[ k = A e^{-\frac{E_a}{RT}} \]

However, for our purposes, we will use the logarithmic form of the Arrhenius equation for two different temperatures:

\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Where:
- \(k_1\) is the rate constant at temperature \(T_1\)
- \(k_2\) is the rate constant at temperature \(T_2\)
- \(E_a\) is the activation energy
- \(R\) is the universal gas constant (8.314 J/mol·K)
- \(T_1\) and \(T_2\) are the absolute temperatures in Kelvin

### Step-by-step Solution:

1. **Identify the given values:**
- \(k_1 = 4.5 \times 10^3 \, s^{-1}\)
- \(k_2 = 1.5 \times 10^4 \, s^{-1}\)
- \(T_1 = 10^\circ C = 10 + 273 = 283 \, K\)
- \(E_a = 60 \, kJ/mol = 60 \times 1000 \, J/mol = 60000 \, J/mol\)
- \(R = 8.314 \, J/mol \cdot K\)

2. **Substitute the values into the Arrhenius equation:**
\[ \log \frac{1.5 \times 10^4}{4.5 \times 10^3} = \frac{60000}{2.303 \times 8.314} \left( \frac{1}{283} - \frac{1}{T_2} \right) \]

3. **Calculate the left side:**
\[ \log \frac{1.5 \times 10^4}{4.5 \times 10^3} = \log \frac{1.5}{4.5} + \log 10 = \log \frac{1.5}{4.5} + 1 \]
\[ \log \frac{1.5}{4.5} = \log \frac{1}{3} \approx -0.477 \]
Therefore,
\[ \log \frac{1.5 \times 10^4}{4.5 \times 10^3} \approx -0.477 + 1 = 0.523 \]

4. **Calculate the right side:**
\[ \frac{60000}{2.303 \times 8.314} \approx \frac{60000}{19.099} \approx 3132.86 \]

5. **Set up the equation:**
\[ 0.523 = 3132.86 \left( \frac{1}{283} - \frac{1}{T_2} \right) \]

6. **Rearranging the equation:**
\[ \frac{1}{T_2} = \frac{1}{283} - \frac{0.523}{3132.86} \]

7. **Calculate \(\frac{0.523}{3132.86}\):**
\[ \frac{0.523}{3132.86} \approx 0.000167 \]

8. **Calculate \(\frac{1}{T_2}\):**
\[ \frac{1}{T_2} \approx \frac{1}{283} - 0.000167 \approx 0.003535 - 0.000167 = 0.003368 \]

9. **Calculate \(T_2\):**
\[ T_2 \approx \frac{1}{0.003368} \approx 297.2 \, K \]

10. **Convert \(T_2\) to Celsius:**
\[ T_2 = 297.2 - 273 \approx 24.2^\circ C \]

### Final Answer:
The temperature at which \(k\) would be \(1.5 \times 10^4 \, s^{-1}\) is approximately \(24.2^\circ C\).