Question

For the reaction PCI₅ → PCI₃ + Cl₂, rate and rate constant are 1.02×10^-4 mol L^-1s^-1 and 3.4×10^-5s^-1 respectively at a given instant. The molar concentration of PCI₅, at that instant is:

• A. 8.0 molL^-1
• B. 3.0 molL^-1
• C. 0.2 molL^-1
• D. 2.0 molL^-1

Answer 1

The Correct Option is B

The given reaction is:

$$\text{PCI}_5 \rightarrow \text{PCI}_3 + \text{Cl}_2$$

We are given the following information:

• Rate constant, $k = 1.02 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}$
• Rate of reaction, $\text{rate} = 3.4 \times 10^{-5} \, \text{s}^{-1}$

Step 1: Use the rate law for the reaction

For the reaction $\text{PCI}_5 \rightarrow \text{PCI}_3 + \text{Cl}_2$, the rate law can be written as:

$$\text{rate} = k \times [\text{PCI}_5]$$

Where $[\text{PCI}_5]$ is the molar concentration of $\text{PCI}_5$, which we need to calculate.

Step 2: Rearrange the rate law to solve for $[\text{PCI}_5]$

$$[\text{PCI}_5] = \frac{\text{rate}}{k}$$

Substitute the given values:

$$[\text{PCI}_5] = \frac{3.4 \times 10^{-5} \, \text{s}^{-1}}{1.02 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}}$$

Step 3: Calculate the concentration of $\text{PCI}_5$

$$[\text{PCI}_5] = 3.03 \, \text{mol L}^{-1}$$

Conclusion:

The molar concentration of $\text{PCI}_5$ at that instant is approximately 3.0 mol L$^{-1}$, so the correct answer is (B) 3.0 mol L$^{-1}$.

Answer 2

The rate of a reaction is given by: $$\text{Rate} = k[\text{PC}_1]$$ Substitute the given values: $$1.02 \times 10^{-4} = (3.4 \times 10^{-5}) \times [\text{PC}_1]$$ Solve for $[\text{PC}_1]$: $$[\text{PC}_1] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}} = 3.0 \, \text{mol L}^{-1}$$