The instantaneous rate of disappearance of $$MnO_4^-$$ ion in the following reaction is $$4.56 \times 10^{-3}$$ Ms$$^{-1}$$
$$2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$$
The rate of appearance of $$I_2$$ is :

The given reaction is:

$$2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$$

The instantaneous rate of disappearance of $$MnO_4^-$$ is given as $$4.56 \times 10^{-3}$$ Ms$$^{-1}$$. This means $$-\frac{d[MnO_4^-]}{dt} = 4.56 \times 10^{-3}$$ Ms$$^{-1}$$.

To find the rate of appearance of $$I_2$$, which is $$\frac{d[I_2]}{dt}$$, we use the stoichiometric coefficients from the balanced equation. The rate of reaction can be expressed in terms of any reactant or product. For the reaction, the rate $$r$$ is defined as:

$$r = -\frac{1}{2} \frac{d[MnO_4^-]}{dt} = \frac{1}{5} \frac{d[I_2]}{dt}$$

We know $$-\frac{d[MnO_4^-]}{dt} = 4.56 \times 10^{-3}$$ Ms$$^{-1}$$. Substituting this into the expression for the rate:

$$r = -\frac{1}{2} \times (-\frac{d[MnO_4^-]}{dt}) = \frac{1}{2} \times (4.56 \times 10^{-3})$$

Calculate the value:

$$\frac{1}{2} \times 4.56 \times 10^{-3} = 2.28 \times 10^{-3}$$ Ms$$^{-1}$$

So, $$r = 2.28 \times 10^{-3}$$ Ms$$^{-1}$$.

Now, using the rate expression for $$I_2$$:

$$r = \frac{1}{5} \frac{d[I_2]}{dt}$$

Substitute the value of $$r$$:

$$2.28 \times 10^{-3} = \frac{1}{5} \frac{d[I_2]}{dt}$$

To solve for $$\frac{d[I_2]}{dt}$$, multiply both sides by 5:

$$\frac{d[I_2]}{dt} = 5 \times 2.28 \times 10^{-3}$$

Calculate the product:

$$5 \times 2.28 = 11.4$$

So, $$\frac{d[I_2]}{dt} = 11.4 \times 10^{-3} = 1.14 \times 10^{-2}$$ Ms$$^{-1}$$.

Therefore, the rate of appearance of $$I_2$$ is $$1.14 \times 10^{-2}$$ Ms$$^{-1}$$.

Comparing with the options:

A. $$4.56 \times 10^{-4}$$ Ms$$^{-1}$$

B. $$1.14 \times 10^{-2}$$ Ms$$^{-1}$$

C. $$1.14 \times 10^{-3}$$ Ms$$^{-1}$$

D. $$5.7 \times 10^{-3}$$ Ms$$^{-1}$$

Hence, the correct answer is Option B.