PCl5(g) - PCl3(g) + Cl2(g)
In the above first order reaction, the concentration of PCl5 reduces fro

Q: Why do we use $2.303$ in the rate constant formula?

The factor $2.303$ is used to convert the natural logarithm ($\ln$) to the common base-10 logarithm ($\log_{10}$). The original formula is $k = \frac{1}{t} \ln\left(\frac{[A_0]}{[A_t]}\right)$.

Q: Does the volume of the container matter in this calculation?

No, because the values are already given in terms of concentration ($\text{mol L}^{-1}$). The volume is already factored into these values.

Q: What if the reaction was second order instead of first order?

If it were a second-order reaction, the integrated rate law would be different: $\frac{1}{[A_t]} - \frac{1}{[A_0]} = kt$. Always check the reaction order first!

Q: Why did we round $1.3413$ to $1$?

The question specifically asked for the value of $x$ as the 'Nearest integer'. Since $1.3413$ is closer to $1$ than to $2$, we round it down to $1$.

Q: Are the units of $k$ always $\text{min}^{-1}$?

For a first-order reaction, the unit of $k$ is always $\text{time}^{-1}$. It could be $\text{s}^{-1}$, $\text{min}^{-1}$, or $\text{hr}^{-1}$ depending on the unit of time used in the calculation.

Question

$	ext{PCl}_5(g) \longrightarrow 	ext{PCl}_3(g) + 	ext{Cl}_2(g)$
In the above first order reaction, the concentration of $	ext{PCl}_5$ reduces from initial concentration $50 	ext{ mol L}^{-1}$ to $10 	ext{ mol L}^{-1}$ in $120$ minutes at $300 	ext{ K}$. The rate constant for the reaction at $300 	ext{ K}$ is $x 	imes 10^{-2} 	ext{ min}^{-1}$. The value of $x$ is ...... . [Given, $\log 5 = 0.6989$]

Sigma Percentile · Accepted Answer

The decomposition of phosphorus pentachloride ($PCl_5$) into phosphorus trichloride ($PCl_3$) and chlorine gas ($Cl_2$) is a classic example of a first-order reaction. In this problem, we are tasked with finding the rate constant $k$ given the initial and final concentrations over a specific time period.

### Analyzing the Setup

Imagine a closed vessel at $300 	ext{ K}$ filled with $PCl_5$ gas. Initially, the concentration is quite high at $50 	ext{ mol L}^{-1}$. As time ticks by, the molecules decompose, and after $120 	ext{ minutes}$, the concentration drops to $10 	ext{ mol L}^{-1}$.

Because the problem explicitly states that this is a **first-order reaction**, we know that the rate of decomposition depends linearly on the concentration of $PCl_5$ present at any given moment. This exponential decay is the hallmark of first-order kinetics.

### The Master Equation

To find the rate constant $k$, we rely on the integrated rate law for a first-order reaction:

$$k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A_t]} ight)$$

Here, $[A_0]$ is the initial concentration, $[A_t]$ is the concentration at time $t$, and the factor $2.303$ is used to convert the natural logarithm ($\ln$) to the base-10 logarithm ($\log_{10}$).

### Substituting and Simplifying

Let's plug our known values into the master equation. We have $t = 120 	ext{ min}$, $[A_0] = 50 	ext{ M}$, and $[A_t] = 10 	ext{ M}$:

$$k = \frac{2.303}{120} \log \left( \frac{50}{10} ight)$$

The fraction inside the logarithm simplifies beautifully:

$$k = \frac{2.303}{120} \log (5)$$

The problem generously provides the value of $\log 5 = 0.6989$. Substituting this in, we get:

$$k = \frac{2.303}{120} 	imes 0.6989$$

### Final Calculation

Now, it's just a matter of careful arithmetic. Multiplying the numerator and dividing by $120$ yields:

$$k = 0.013413 	ext{ min}^{-1}$$

The question asks for the answer in the specific format of $x 	imes 10^{-2} 	ext{ min}^{-1}$. To match this, we shift the decimal point two places to the right:

$$k = 1.3413 	imes 10^{-2} 	ext{ min}^{-1}$$

Comparing this to the required format, we find that $x = 1.3413$. Since we need to round off to the **nearest integer**, our final answer is simply $1$.

Enter Numerical Value:

Visualized Solution

$Reaction Setup$

$First Order Rate Law$

$Substituting Values$

$Simplifying the Log Term$

$Calculating k$

$Formatting the Answer$

$Food for Thought$

The Sigma Insight: Order and Molecularity

Analyzing the Setup

The Master Equation

Substituting and Simplifying

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Order Rate Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Log Term

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating k

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formatting the Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought