The first order rate constant for the decomposition of CaCO$$_3$$ at 700 K is $$6.36 \times 10^{-3}$$ s$$^{-1}$$ and activation energy is 209 kJ mol$$^{-1}$$. Its rate constant (in s$$^{-1}$$) at 600 K is $$x \times 10^{-6}$$. The value of x is _________. (Nearest integer)
[Given R = 8.31 J K$$^{-1}$$ mol$$^{-1}$$; log $$6.36 \times 10^{-3}$$ = -2.19, $$10^{-4.79}$$ = $$1.62 \times 10^{-5}$$]

We have to compare the rate constants at two temperatures. The Arrhenius equation in its logarithmic (base 10) form is first written:

$$\log_{10}\!\left(\frac{k_2}{k_1}\right)= -\frac{E_a}{2.303\,R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

Here, $$k_1$$ is the rate constant at $$T_1$$, $$k_2$$ is the rate constant at $$T_2$$, $$E_a$$ is the activation energy and $$R$$ is the gas constant.

Given data are

$$k_1 = 6.36 \times 10^{-3}\ \text{s}^{-1},\ \ T_1 = 700\ \text{K},\ \ T_2 = 600\ \text{K},\ \ E_a = 209\ \text{kJ mol}^{-1} = 209000\ \text{J mol}^{-1},\ \ R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}$$.

First, compute the reciprocal-temperature difference:

$$\frac{1}{T_2}-\frac{1}{T_1}= \frac{1}{600}-\frac{1}{700} = \frac{700-600}{600 \times 700} = \frac{100}{420000}= \frac{1}{4200}=0.000238095\ \text{K}^{-1}.$$

Next, evaluate the coefficient $$\dfrac{E_a}{2.303\,R}$$:

$$\frac{E_a}{2.303\,R}= \frac{209000}{2.303 \times 8.31} = \frac{209000}{19.144}\approx 1.0918 \times 10^{4}.$$

Multiply this coefficient by the temperature difference and insert the minus sign dictated by the formula:

$$-\frac{E_a}{2.303\,R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)= -\left(1.0918 \times 10^{4}\right)\left(0.000238095\right) \approx -2.598.$$

So

$$\log_{10}\!\left(\frac{k_2}{k_1}\right)= -2.598.$$

This relation means

$$\log_{10}k_2 = \log_{10}k_1 - 2.598.$$

We already know $$\log_{10}k_1 = \log_{10}(6.36 \times 10^{-3}) = -2.19$$ (value supplied). Substituting:

$$\log_{10}k_2 = -2.19 - 2.598 = -4.788.$$

Changing the logarithm back to the numerical value, we recall the given approximation $$10^{-4.79} = 1.62 \times 10^{-5}$$. Since $$-4.788$$ is virtually $$-4.79$$, we obtain

$$k_2 \approx 1.62 \times 10^{-5}\ \text{s}^{-1}.$$

The question expresses $$k_2$$ in the form $$x \times 10^{-6}\ \text{s}^{-1}$$. Converting:

$$1.62 \times 10^{-5}\ = 16.2 \times 10^{-6}.$$

The nearest integer to $$16.2$$ is $$16$$.

So, the answer is $$16$$.