The activation energy of one of the reactions in a biochemical process is 532611 J mol$$^{-1}$$. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k$$_{300}$$ = x $$\times$$ 10$$^{-3}$$k$$_{310}$$. The value of x is ______.
[Given: ln 10 = 2.3 R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$]

We use the Arrhenius equation in logarithmic form:

$$\ln\left(\frac{k_{300}}{k_{310}}\right) = -\frac{E_a}{R}\left(\frac{1}{300} - \frac{1}{310}\right)$$

Compute the temperature term:

$$\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000}$$

Substitute the values ($$E_a = 532611$$ J mol$$^{-1}$$, $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$):

$$\ln\left(\frac{k_{300}}{k_{310}}\right) = -\frac{532611}{8.3} \times \frac{10}{93000}$$

$$= -\frac{5326110}{771900} = -6.9$$

Since $$\ln 10 = 2.3$$, we have $$-6.9 = -3 \times 2.3 = -3 \ln 10 = \ln(10^{-3})$$.

Therefore:

$$\frac{k_{300}}{k_{310}} = 10^{-3}$$

$$k_{300} = 10^{-3} \times k_{310} = 1 \times 10^{-3} \times k_{310}$$

Comparing with $$k_{300} = x \times 10^{-3} k_{310}$$, we get $$x = 1$$.