Decomposition of $$H_2O_2$$ follows a first order reaction. In fifty minutes the concentration of $$H_2O_2$$ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of $$H_2O_2$$ reaches 0.05 M, the rate of formation of $$O_2$$ will be:

We are told that the decomposition of $$H_2O_2$$ follows first-order kinetics. For any first-order reaction the integrated rate law is stated first:

$$\ln \frac{[A]_0}{[A]} = kt.$$

If we prefer common logarithms, the same relation is written as

$$k = \frac{2.303}{t}\,\log\!\frac{[A]_0}{[A]}.$$

We have an initial concentration $$[A]_0 = 0.5\;{\rm M}$$ that falls to $$[A] = 0.125\;{\rm M}$$ in a time $$t = 50\;{\rm min}.$$ Substituting these numbers gives

$$k = \frac{2.303}{50}\,\log\!\frac{0.5}{0.125}.$$

The ratio inside the logarithm is $$\dfrac{0.5}{0.125}=4,$$ and $$\log 4 = 0.6020.$$ Hence

$$k = \frac{2.303 \times 0.6020}{50}.$$

Multiplying the numerator we obtain $$2.303 \times 0.6020 = 1.386,$$ so

$$k = \frac{1.386}{50} = 0.0277\;{\rm min^{-1}}.$$

For a first-order reaction the instantaneous rate of disappearance of $$H_2O_2$$ is

$$-\frac{d[H_2O_2]}{dt} = k[H_2O_2].$$

Now we are interested in the moment when $$[H_2O_2] = 0.05\;{\rm M}.$$ Putting this value into the expression for the rate, we get

$$-\frac{d[H_2O_2]}{dt} = 0.0277 \times 0.05.$$

Carrying out the multiplication,

$$-\frac{d[H_2O_2]}{dt} = 1.385 \times 10^{-3}\;{\rm mol\;L^{-1}\,min^{-1}}.$$

This is the rate at which $$H_2O_2$$ is being consumed. The balanced chemical equation for the decomposition is

$$2H_2O_2 \rightarrow 2H_2O + O_2.$$

From the stoichiometry we see that two moles of $$H_2O_2$$ produce one mole of $$O_2$$, so the rate of formation of oxygen is one-half of the rate of disappearance of $$H_2O_2$$. Thus,

$$\frac{d[O_2]}{dt} = \frac{1}{2}\left(-\frac{d[H_2O_2]}{dt}\right) = \frac{1}{2}\,(1.385 \times 10^{-3}).$$

Dividing by two,

$$\frac{d[O_2]}{dt} = 6.93 \times 10^{-4}\;{\rm mol\;L^{-1}\,min^{-1}}.$$

Because the options are expressed simply in $$\rm mol\;min^{-1},$$ and the numerical value matches exactly, we choose the corresponding option.

Hence, the correct answer is Option D.