The reaction X→ products is a first order reaction. In 40 minutes, the concentration of X changes from 1.0M to 0.25M. What is the initial rate of reaction when [X]=0.1M ? (log4=0.60)

🧩 Key Concepts Needed

1. First-Order Integrated Rate Law
   For a reaction $X \rightarrow \text{products}$,
   $\ln\left(\frac{[X]_0}{[X]}\right) = k t$
   or, in base-10 logarithm,
   $k = \frac{2.303}{t}\;\log\left(\frac{[X]_0}{[X]}\right)$
   where $k$ is the rate constant.

2. Initial Rate for First Order
   The instantaneous (initial) rate when concentration is $[X]$ is
   $\text{Rate} = k\,[X]$

🔍 Logical Chain of Thought

1. Determine $k$
   • We know $[X]_0 = 1.0\,\text{M}$, $[X] = 0.25\,\text{M}$, $t = 40\,\text{min}$.
   • Plug into the integrated law to solve $k$.
2. Plug New Concentration
   • Once $k$ is known, use $[X]=0.10\,\text{M}$ in $\text{Rate} = k[X]$.
3. Compute & Report Units
   • For first order, $k$ has units $\text{time}^{-1}$ (here $\text{min}^{-1}$).
   • Rate will have units $\text{M}\,\text{min}^{-1}$.

That sequence gives the required initial rate.

🌟 Imagine a Melting Ice Cube Game!

1. Ice Cube = Reactant X
   Think of the reactant $X$ as an ice cube that is slowly melting.
2. Stop-Watch = Time
   You start a timer to see how long it takes for the cube to shrink to certain sizes.
3. Rule of the Game (First Order)
   Every bit of ice that remains decides how fast the next bit melts. So the less ice you have, the slower it melts—just like a first-order reaction!
4. What We Observed
   In 40 minutes, the cube shrank from a big size (1.0 M) to a quarter of that (0.25 M).
5. What We Want
   If you now start with a tiny cube (0.1 M), how fast will it melt right at the beginning?

In science words, we will:

• Use the first-order formula to find the ‘melting speed constant’ ($k$).
• Multiply that $k$ by the new small size (0.1 M) to get the initial speed of melting.

That’s all!

Step 1: Calculate the Rate Constant $k$

Given:
$[X]_0 = 1.0\,\text{M}$
$[X] = 0.25\,\text{M}$
$t = 40\,\text{min}$

For a first-order reaction,

$k = \frac{2.303}{t}\,\log\left(\frac{[X]_0}{[X]}\right)$

Substitute the values:

$k = \frac{2.303}{40}\;\log\left(\frac{1.0}{0.25}\right)$

Since $\dfrac{1.0}{0.25}=4$ and $\log 4 = 0.60$ (given),

$k = \frac{2.303}{40} \times 0.60$

$k = \frac{1.3818}{40}$

$k = 0.0345\,\text{min}^{-1}$

Step 2: Find the Initial Rate at $[X] = 0.10\,\text{M}$

For first order,

$\text{Rate} = k [X]$

$\text{Rate} = 0.0345\,\text{min}^{-1} \times 0.10\,\text{M}$

$\text{Rate} = 0.00345\,\text{M} \; \text{min}^{-1}$

Final Answer

$\boxed{\text{Initial rate} \; \approx 3.5 \times 10^{-3}\; \text{M min}^{-1}}$