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Step-by-Step Solution
Step 1: Identify the relevant integrated rate law for a first-order reaction
For a first-order reaction, the integrated rate law is given by
k = \frac{2.303}{t} \log \left(\frac{[A_0]}{[A_t]}\right) ,
where
[A_0] is the initial concentration,
[A_t] is the concentration at time t, and
k is the rate constant.
Step 2: Substitute the given data
Given:
• Decomposition: PCl_5(g) \longrightarrow PCl_3(g) + Cl_2(g)
• Initial concentration [A_0] = 50\,\text{mol L}^{-1}
• Concentration after t = 120\,\text{min} is [A_t] = 10\,\text{mol L}^{-1}
• The reaction is first-order at 300\,K
• \log(5) = 0.6989
Step 3: Evaluate the logarithmic term
The ratio \frac{[A_0]}{[A_t]} = \frac{50}{10} = 5 .
Hence,
\log \left(\frac{50}{10}\right) = \log(5) = 0.6989.
Step 4: Calculate the rate constant, k
Substitute values into the first-order rate law:
k = \frac{2.303}{120}\,\log\left(\frac{50}{10}\right) = \frac{2.303}{120}\times 0.6989.
First compute the product of 2.303 and 0.6989:
2.303 \times 0.6989 \approx 1.609.
Now divide by 120:
k \approx \frac{1.609}{120} \approx 0.0134\,\text{min}^{-1}.
Thus, k \approx 1.34 \times 10^{-2}\,\text{min}^{-1}.
Step 5: Extract the value of x
The problem states that k = x \times 10^{-2}\,\text{min}^{-1}.
From our calculation, k \approx 1.34 \times 10^{-2}\,\text{min}^{-1},
so x \approx 1.34.
If one needs the nearest integer for x, it would be 1.
