1. Use the Arrhenius equation:

$k = A e^{-E_a/(RT)}$

Taking natural log:

$\ln k = \ln A - \frac{E_a}{R}\cdot \frac{1}{T}$

This equation will help us test all statements.

2. Check statement A:

$k = A e^{-E_a/(RT)}$

For fixed $A$ and $T$, if $E_a$ increases, then $e^{-E_a/(RT)}$ decreases, so $k$ decreases.

So, A is correct.

3. Check statement B:

Temperature coefficient means the factor by which rate constant increases for a $10^\circ C$ rise in temperature:

$\text{temperature coefficient} = \frac{k_{T+10}}{k_T}$

Using Arrhenius form,

$\ln\left(\frac{k_{T+10}}{k_T}\right) = \frac{E_a}{R}\left(\frac{1}{T} - \frac{1}{T+10}\right)$

This increases with $E_a$. Hence, larger activation energy gives a larger temperature coefficient.

So, B is correct.

4. Check statement C:

From Arrhenius equation,

$\ln k = \ln A - \frac{E_a}{RT}$

The sensitivity of $k$ to temperature is larger at low $T$. More precisely,

$\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$

Since this is proportional to $\frac{1}{T^2}$, the effect of temperature increase is greater at lower temperature than at higher temperature.

So, C is correct.

5. Check statement D:

From

$\ln k = \ln A - \frac{E_a}{R}\cdot \frac{1}{T}$

A plot of $\ln k$ versus $\frac{1}{T}$ is a straight line with slope

$-\frac{E_a}{R}$

So, D is correct.

6. Count correct statements:

A, B, C, and D are all correct.

Therefore, the number of correct statements is

$4$

7. Comparison with stored answer:

Stored correct answer = $3$

Derived answer = $4$

Hence, I disagree with the stored answer. The stored answer likely missed statement C, which is true because the temperature dependence of $k$ is stronger at lower temperatures.