The reaction `CH_(3) – CH_(2) – NO_(2) + OH^(–) rarr CH_(3) – CH – NO_(2) + H_(2)O` obeys the rate law for pseudo first order kinetics in the presence of a large excess of hydroxide ion. If 1% of nitro ethane undergoes reaction in half a minute when the reactant concentration is 0.002 M, What is the pseudo first order rate constant ?

Options

A. `0.02" min"^(-1)`
B. `0.05" min"^(-1)`
C. `0.01" min"^(-1)`
D. `0.04" min"^(-1)`

Answer

To solve the problem, we need to find the pseudo first-order rate constant (k) for the reaction given the information provided. Here’s a step-by-step solution:

### Step 1: Understand the Reaction and Given Data
The reaction is:
\[ \text{CH}_3\text{CH}_2\text{NO}_2 + \text{OH}^- \rightarrow \text{CH}_3\text{CH}\text{NO}_2 + \text{H}_2\text{O} \]

We are given:
- The concentration of nitroethane, \([A_0] = 0.002 \, \text{M}\)
- Time, \(t = 0.5 \, \text{minutes}\)
- 1% of nitroethane reacts.

### Step 2: Calculate the Amount Reacted (x)
1% of the initial concentration of nitroethane reacts:
\[ x = 1\% \text{ of } 0.002 \, \text{M} = \frac{1}{100} \times 0.002 = 0.00002 \, \text{M} \]

### Step 3: Calculate the Final Concentration
The final concentration after the reaction will be:
\[ [A] = [A_0] - x = 0.002 \, \text{M} - 0.00002 \, \text{M} = 0.00198 \, \text{M} \]

### Step 4: Use the First-Order Rate Equation
For a first-order reaction, the rate constant \(k\) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \]

### Step 5: Substitute the Values into the Equation
Substituting the values we have:
- \( t = 0.5 \, \text{minutes} = 0.5 \)
- \( [A_0] = 0.002 \, \text{M} \)
- \( [A] = 0.00198 \, \text{M} \)

Now substituting into the equation:
\[ k = \frac{2.303}{0.5} \log \left( \frac{0.002}{0.00198} \right) \]

### Step 6: Calculate the Logarithm
Calculate the fraction:
\[ \frac{0.002}{0.00198} = 1.01010101 \]

Now find the logarithm:
\[ \log(1.01010101) \approx 0.004321 \]

### Step 7: Calculate k
Now substitute back into the equation for k:
\[ k = \frac{2.303}{0.5} \times 0.004321 \]
\[ k = 4.606 \times 0.004321 \]
\[ k \approx 0.0199 \, \text{min}^{-1} \]

### Step 8: Round the Value
Rounding to two decimal places, we find:
\[ k \approx 0.02 \, \text{min}^{-1} \]

### Final Answer
The pseudo first-order rate constant \( k \) is approximately \( 0.02 \, \text{min}^{-1} \).

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