© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Known Quantities
• Rate constant at 700 K, K_{700} = 6.36 \times 10^{-3}\,\text{s}^{-1}
• Rate constant at 600 K, K_{600} = x \times 10^{-6}\,\text{s}^{-1}
• Activation energy, E_{a} = 209\,\text{kJ mol}^{-1} = 209\,000\,\text{J mol}^{-1}
• Universal gas constant, R = 8.31\,\text{J K}^{-1}\text{mol}^{-1}
• Temperatures: T_{1} = 700\,\text{K} , T_{2} = 600\,\text{K} .
Step 2: Write the Arrhenius Relation in Logarithmic Form
The Arrhenius equation connecting two rate constants K_{T_1} and K_{T_2} is given by:
\log \left(\frac{K_{T_2}}{K_{T_1}}\right)
= -\frac{E_a}{2.303 \, R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right).
Step 3: Substitute the Known Values
We want the ratio K_{600}/K_{700} . Rearranging and using the given data:
\log \left(\frac{K_{600}}{K_{700}}\right)
= -\frac{209000}{2.303 \times 8.31}
\left(\frac{1}{600} - \frac{1}{700}\right).
However, the solution as provided in the reference rearranges it to:
\log \left(\frac{K_{700}}{K_{600}}\right)
= \frac{209000}{2.303 \times 8.31}
\left(\frac{1}{600} - \frac{1}{700}\right).
Either form leads to the same final value for K_{600} .
Step 4: Use the Logarithm Values
From the given solution steps, we have:
\displaystyle
\log\left(\frac{K_{700}}{K_{600}}\right)
= \log\left(\frac{6.36 \times 10^{-3}}{K_{600}}\right) = 2.6.
Also, we are given \log(6.36 \times 10^{-3}) = -2.19.
Step 5: Solve for K_{600}
From the expression above:
\log\left(\frac{6.36 \times 10^{-3}}{K_{600}}\right)
= -2.19 - (\log K_{600}) = 2.6.
Hence,
-2.19 - \log(K_{600}) = 2.6
\quad\Longrightarrow\quad \log(K_{600}) = -2.19 - 2.6 = -4.79.
Therefore, we take the antilog:
K_{600} = 10^{-4.79} \approx 1.62 \times 10^{-5}\,\text{s}^{-1}.
Step 6: Express in the Given Form
We need K_{600} = x \times 10^{-6}\,\text{s}^{-1}. Since 1.62 \times 10^{-5} = 16.2 \times 10^{-6} ,
the value of x (nearest integer) is:
\displaystyle x = 16.
Final Answer
\boxed{16}
