PCl$$_5$$(g) $$\to$$ PCl$$_3$$(g) + Cl$$_2$$(g)
In the above first order reaction the concentration of PCl$$_5$$ reduces from initial concentration 50 mol L$$^{-1}$$ to 10 mol L$$^{-1}$$ in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x $$\times 10^{-2}$$ min$$^{-1}$$. The value of x is ___.
[Given log 5 = 0.6989]

For a first-order reaction, the rate constant is given by:

$$k = \frac{2.303}{t} \log\frac{[\text{A}]_0}{[\text{A}]}$$

Given: $$[\text{PCl}_5]_0 = 50$$ mol L$$^{-1}$$, $$[\text{PCl}_5] = 10$$ mol L$$^{-1}$$, and $$t = 120$$ min.

$$k = \frac{2.303}{120} \times \log\frac{50}{10} = \frac{2.303}{120} \times \log 5$$

Using $$\log 5 = 0.6989$$:

$$k = \frac{2.303 \times 0.6989}{120} = \frac{1.6094}{120} = 0.01341 \text{ min}^{-1}$$

$$k = 1.34 \times 10^{-2} \text{ min}^{-1}$$

Since $$k = x \times 10^{-2}$$ min$$^{-1}$$, we have $$x \approx 1.34$$, which rounds to the nearest integer as $$\mathbf{1}$$.