The reaction between `Cr_(2)O_(7)^(2-)` and `HNO_(2)` in an acidic medium is
`Cr_(2)O_(7)^(2-) + 5H^(o+) + 3HNO_(2) rarr 2Cr^(3+) + 3NO_(3)^(ɵ) + 4H_(2)O`.
The rate of disappearance of `Cr_(2)O_(7)^(2-)` is found to be `2.4 xx 10^(-4) mol L^(-1) s^(-1)` during measured time interval. What will be the rate of disappearance of `HNO_(2)` during the same time interval?
(a) `2.4xx10^(-4)`
(b) `7.2xx10^(-4)`
(c) `4.8xx10^(-4)`
(d) `0.8xx10^(-4)`

Options

A. `2.4xx10^(-4)`
B. `7.2xx10^(-4)`
C. `4.8xx10^(-4)`
D. `0.8xx10^(-4)`

Explanation

To solve the problem, we need to determine the rate of disappearance of \( HNO_2 \) based on the given rate of disappearance of \( Cr_2O_7^{2-} \).

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction is given as:
\[ Cr_2O_7^{2-} + 5H^+ + 3HNO_2 \rightarrow 2Cr^{3+} + 3NO_3^{-} + 4H_2O \]

2. **Identify the stoichiometric coefficients**:
From the balanced equation, we can see the stoichiometric coefficients:
- For \( Cr_2O_7^{2-} \): 1
- For \( HNO_2 \): 3

3. **Set up the rate expressions**:
The rate of disappearance of a reactant can be expressed in terms of its stoichiometric coefficient:
\[ -\frac{1}{1} \frac{d[Cr_2O_7^{2-}]}{dt} = -\frac{1}{3} \frac{d[HNO_2]}{dt} \]

4. **Substitute the known rate**:
We know the rate of disappearance of \( Cr_2O_7^{2-} \) is given as:
\[ \frac{d[Cr_2O_7^{2-}]}{dt} = 2.4 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]
Substituting this value into the rate expression gives:
\[ -\frac{d[Cr_2O_7^{2-}]}{dt} = 2.4 \times 10^{-4} \]

5. **Rearranging the rate expression**:
Using the rate expression we derived:
\[ 2.4 \times 10^{-4} = \frac{1}{3} \frac{d[HNO_2]}{dt} \]
Rearranging to find the rate of disappearance of \( HNO_2 \):
\[ \frac{d[HNO_2]}{dt} = 3 \times 2.4 \times 10^{-4} \]

6. **Calculate the rate of disappearance of \( HNO_2 \)**:
\[ \frac{d[HNO_2]}{dt} = 7.2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Conclusion:
The rate of disappearance of \( HNO_2 \) during the same time interval is:
\[ \boxed{7.2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}} \]