The reaction `S_(2)O_(8)^(2-) + 3I^(ɵ) rarr 2SO_(4)^(2-) + I_(3)^(ɵ)` is of first order both with respect to persulphate and iofide ions. Taking the initial concentration as `a` and `b`, respectively, and taking `x` as the concentration of the triofide at time `t`, a differential rate equation can be written.
Two suggested mechanism for the reaction are:
I. `S_(2)O_(8)^(2-)+I^(ɵ) hArr SO_(4)I^(ɵ)+SO_(4)^(2-) ("fast")`
`I^(ɵ)+SO_(4)I^(ɵ) overset(k_(1))rarrI_(2) + SO_(4)^(2-)` (show)
`I^(ɵ) + I_(2) overset(k_(2))rarr I_(3)^(ɵ) ("fast")`
II. `S_(2)O_(8)^(2-) + I^(ɵ) overset(k_(1))rarr S_(2)O_(8) I^(2-) (slow)`
`S_(2)O_(8)I^(3-) overset(k_(2))rarr2SO_(4)^(2-)+I^(o+) ("fast")`
`I^(o+) + I^(ɵ) overset(k_(3)) rarr I_(2) ("fast")`
`I_(2) + I^(o+) overset(k_(4))rarr I_(3)^(ɵ) ("fast")`
The general difference equation for the above reaction is

Options

A. `(dx)/(dt) = k[a-x][b-3x] (kgt0)`
B. `(dx)/(dt) = -k[a-x][b-3x] (kgt0)`
C. `(dx)/(dt) = k[a-x][b-x] (kgt0)`
D. `(dx)/(dt) = -k[a-x][b-x] (kgt0)`

Explanation

To derive the differential rate equation for the given reaction and identify the correct mechanism, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ S_2O_8^{2-} + 3I^{-} \rightarrow 2SO_4^{2-} + I_3^{-} \]

### Step 2: Define initial concentrations
Let:
- The initial concentration of \( S_2O_8^{2-} \) be \( a \).
- The initial concentration of \( I^{-} \) be \( b \).
- The concentration of \( I_3^{-} \) formed at time \( t \) be \( x \).

### Step 3: Write the concentrations at time \( t \)
At time \( t \):
- The concentration of \( S_2O_8^{2-} \) will be \( a - x \).
- The concentration of \( I^{-} \) will be \( b - 3x \).
- The concentration of \( SO_4^{2-} \) formed will be \( 2x \).
- The concentration of \( I_3^{-} \) formed will be \( x \).

### Step 4: Write the rate of the reaction
The rate of the reaction can be expressed as:
\[ \text{Rate} = -\frac{d[S_2O_8^{2-}]}{dt} = -\frac{d[I^{-}]}{dt} = \frac{d[I_3^{-}]}{dt} \]

### Step 5: Express the rate in terms of concentrations
Since the reaction is first order with respect to both \( S_2O_8^{2-} \) and \( I^{-} \), we can write the rate law as:
\[ \text{Rate} = k [S_2O_8^{2-}] [I^{-}] \]
Substituting the concentrations:
\[ \text{Rate} = k (a - x)(b - 3x) \]

### Step 6: Write the differential equation
Since the rate is equal to \( -\frac{dx}{dt} \):
\[ -\frac{dx}{dt} = k (a - x)(b - 3x) \]

### Step 7: Rearranging the equation
This gives us the differential equation:
\[ \frac{dx}{dt} = -k (a - x)(b - 3x) \]

### Conclusion
The general differential equation for the reaction is:
\[ \frac{dx}{dt} = -k (a - x)(b - 3x) \]