Question

For the reaction PCl5​→PCl3​+Cl2​, the rate and rate constant are 1.02×10−4mol L−1s−1 and 3.4×10−5s−1, respectively, at a given instant. The molar concentration of PCl5​ at that instant is:

• A. 8.0mol/L
• B. 3.0mol/L
• C. 0.2mol/L
• D. 2.0mol/L

Hint:

The concentration of PCl5​ can be determined using the rate law Rate=k[PCl5​].

Answer 1

The Correct Option is B

To determine the molar concentration of \(PCl_5\) at the given instant, we need to understand the relationship between the rate of the reaction and the concentration, which for a first-order reaction is given by the formula:

\(r = k \cdot [\text{PCl}_5]\) 

Where:

• \(r\) is the rate of the reaction.
• \(k\) is the rate constant.
• \([\text{PCl}_5]\) is the molar concentration of \(PCl_5\).

Given in the problem:

• \(r = 1.02 \times 10^{-4} \, \text{mol L}^{-1}\text{s}^{-1}\)
• \(k = 3.4 \times 10^{-5} \, \text{s}^{-1}\)

Substitute the given values into the formula:

\(1.02 \times 10^{-4} = 3.4 \times 10^{-5} \times [\text{PCl}_5]\)

To find \([\text{PCl}_5]\), rearrange the equation:

\([\text{PCl}_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}\)

Calculating the above expression:

\([\text{PCl}_5] = \frac{1.02}{3.4} \times \frac{10^{-4}}{10^{-5}} = \frac{1.02}{3.4} \times 10^{1}\)

\([\text{PCl}_5] = 0.3 \times 10 = 3.0 \, \text{mol/L}\)

Therefore, the molar concentration of \(PCl_5\) at that instant is 3.0 mol/L, which matches the correct option among the given choices.