For a reaction, consider the plot of ln k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is $$10^{-5} s^{-1}$$, then the rate constant at 500 K is:

We have a straight-line plot of $$\ln k$$ versus $$1/T$$ for the reaction. For any Arrhenius plot, the governing relation is first written as

$$k \;=\; A\,e^{-\dfrac{E_a}{RT}}.$$

Taking natural logarithms on both sides gives the linear form

$$\ln k \;=\; \ln A \;-\;\dfrac{E_a}{R}\;\left(\dfrac1T\right).$$

In the form $$y = mx + c$$, the ordinate is $$y = \ln k$$, the abscissa is $$x = 1/T$$, and the slope is

$$m = -\dfrac{E_a}{R}.$$

From the figure (reading two well-separated points and calculating), the slope of the straight line is found to be approximately

$$m \;=\; -4.6 \times 10^{3}\;{\rm K}.$$

Stating this numerically,

$$-\dfrac{E_a}{R} = -4.6 \times 10^{3}\;{\rm K}, \qquad\text{so}\qquad \dfrac{E_a}{R} = 4.6 \times 10^{3}\;{\rm K}.$$

Now we wish to compare rate constants at two temperatures. Using the Arrhenius form for two temperatures $$T_1$$ and $$T_2$$, the standard manipulation is

$$\ln\!\left(\dfrac{k_2}{k_1}\right) \;=\; -\dfrac{E_a}{R}\;\Bigl(\dfrac1{T_2} - \dfrac1{T_1}\Bigr).$$

First, list the known data:

$$T_1 = 400\;{\rm K},\qquad k_1 = 10^{-5}\;{\rm s^{-1}},$$

$$T_2 = 500\;{\rm K},\qquad k_2 = \text{?}$$

Calculate the difference in reciprocal temperatures:

$$\dfrac1{T_2} - \dfrac1{T_1} \;=\; \dfrac1{500} \;-\; \dfrac1{400}.$$

Writing both denominators explicitly,

$$\dfrac1{500} = 2.0 \times 10^{-3}\;{\rm K^{-1}},\qquad \dfrac1{400} = 2.5 \times 10^{-3}\;{\rm K^{-1}}.$$

So,

$$\dfrac1{T_2} - \dfrac1{T_1} = (2.0 - 2.5)\times10^{-3} = -0.5 \times 10^{-3} \;=\; -5.0 \times 10^{-4}\;{\rm K^{-1}}.$$

Substituting slope information $$E_a / R = 4.6 \times 10^{3}\;{\rm K}$$ into the two-temperature equation gives

$$\ln\!\left(\dfrac{k_2}{k_1}\right) = -\bigl(4.6 \times 10^{3}\bigr)\;\Bigl(-5.0 \times 10^{-4}\Bigr).$$

Multiplying the two negatives yields a positive result:

$$\ln\!\left(\dfrac{k_2}{k_1}\right) = (4.6 \times 10^{3})\,(5.0 \times 10^{-4}) = 2.3.$$

Recognising that $$\ln 10 \approx 2.303$$, we can rewrite

$$\ln\!\left(\dfrac{k_2}{k_1}\right) \;\approx\; \ln 10.$$

Therefore,

$$\dfrac{k_2}{k_1} \;\approx\; 10.$$

Finally, multiplying by the known $$k_1$$ value,

$$k_2 \;=\; 10\,k_1 \;=\; 10 \times 10^{-5}\;{\rm s^{-1}} \;=\; 10^{-4}\;{\rm s^{-1}}.$$

Hence, the correct answer is Option A.