Why is the rate of disappearance negative? Rate of disappearance is given as $-\frac{\Delta [A]}{\Delta t}$ where $\ce{A}$ is a reactant. However, using this formula, the rate of disappearance cannot be negative.

$\Delta [A]$ will be negative, as $[A]$ will be lower at a later time, since it is being used up in the reaction. Then, $[A]_{\text{final}} - [A]_{\text{initial}}$ will be negative. Therefore, the numerator in $-\frac{\Delta [A]}{\Delta t}$ will be negative.

$\Delta t$ will be positive because final time minus initial time will be positive.

This means that $-\frac{\Delta [A]}{\Delta t}$ will evaluate to $(-)\frac{(-)}{(+)} = (-) \cdot (-) =(+)$

However, we still write the rate of disappearance as a negative number. Also, if you think about it, a negative rate of disappearance is essentially a positive rate of appearance. The reactants disappear at a positive rate, so why isn't the rate of disappearance positive?
Reaction rates are generally by convention given based on the formation of the product, and thus reaction rates are positive. So, for the reaction:

$$\ce{A-&gt;B}$$

$$\text{Rate} = \frac{\Delta[\ce{B}]}{\Delta t}$$

To ensure that you get a positive reaction rate, the rate of disappearance of reactant has a negative sign:

$$\text{Rate} = -\frac{\Delta[\ce{A}]}{\Delta t}=\frac{\Delta[\ce{B}]}{\Delta t}$$