For Reaction Mixture 1 , the student found that 92 seconds were required. On
dividing Equation 5 for Reaction Mixture 1 by Equation 5 for Reaction Mixture
\(4,\) and after canceling out the common terms \(\left(k^{\prime} \text { and
terms in }\left[\mathrm{I}^{-}\right] \text {and
}\left[\mathrm{BrO}_{3}^{-}\right]\right)\), they got the following equation:
$$\frac{10.9}{45}=\left(\frac{0.020}{0.040}\right)^{p}=\left(\frac{1}{2}\right)$$
Recognizing that \(10.9 / 45\) is about equal to one-fourth, they obtained an
approximate value for \(p .\) What was that value?\(p=\)_______.
By taking logarithms of both sides of the equation, the student got an exact
value for \(p\). What was that value?
Since orders of reactions are often integers, the student rounded their value
of \(p\) to the nearest integer and reported that value as the order of the
reaction with respect to \(\mathbf{H}^{+}\). \(p=\)_______.
