For the reaction, $$aA + bB \to cC + dD$$, the plot of log k v/s $$\frac{1}{T}$$ is given below:

The temperature at which the rate constant of the reaction is $$10^{-4}$$ s$$^{-1}$$ is ______ K.
(Rounded-off to the nearest integer) [Given: The rate constant of the reaction is $$10^{-5}$$ s$$^{-1}$$ at 500 K.]

The Arrhenius equation is:

$$\mathrm{\log k = \log A - \frac{E_a}{2.303RT}}$$

Comparing with the straight-line equation:

$$\mathrm{y = mx + c}$$

for a plot of $$\mathrm{\log k}$$ versus $$\mathrm{\frac{1}{T}}$$:

$$\mathrm{Slope = -\frac{E_a}{2.303R}}$$

Given slope:

$$\mathrm{m = -10000\ K}$$

Using the two-point equation:

$$\mathrm{\log k_2 - \log k_1 = m\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$$

Given:

$$\mathrm{k_1 = 10^{-5}\ s^{-1}}$$

$$\mathrm{k_2 = 10^{-4}\ s^{-1}}$$

$$\mathrm{T_1 = 500\ K}$$

Substituting values:

$$\mathrm{\log(10^{-4}) - \log(10^{-5}) = -10000\left(\frac{1}{T_2}-\frac{1}{500}\right)}$$

$$\mathrm{-4 - (-5) = -10000\left(\frac{1}{T_2}-\frac{1}{500}\right)}$$

$$\mathrm{1 = -10000\left(\frac{1}{T_2}-\frac{1}{500}\right)}$$

$$\mathrm{-\frac{1}{10000} = \frac{1}{T_2}-\frac{1}{500}}$$

$$\mathrm{\frac{1}{T_2} = \frac{1}{500}-\frac{1}{10000}}$$

Taking LCM:

$$\mathrm{\frac{1}{T_2} = \frac{20}{10000}-\frac{1}{10000}}$$

$$\mathrm{\frac{1}{T_2} = \frac{19}{10000}}$$

$$\mathrm{T_2 = \frac{10000}{19}}$$

$$\mathrm{T_2 \approx 526.31\ K}$$

Rounding to nearest integer:

$$\boxed{\mathrm{526}}$$