The temperature of a system can be manipulated to increase the reaction rate. The temperature affects the rate of the reaction because as temperature increases, so does the average kinetic energy. This means that molecules (reactants) are moving around more quickly, resulting in more frequent molecular collisions, resulting in an increased rate of product formation.

The average kinetic energy is considered to be a sole function of temperature; pressure is not a factor. All of the other statements are assumptions of kinetic molecular theory. 

Recall that \(\displaystyle r[=]\frac{M}{s}\) and that \(\displaystyle A\) and \(\displaystyle B\) are both molar concentrations. Then,

\(\displaystyle \frac{M}{s} = k(\frac{M}{s})^{1/2}(\frac{M}{s})^{2/3}=k(\frac{M}{s})^{7/6}\)

Dividing both sides by \(\displaystyle (\frac{M}{s})^{7/6}\) gives

\(\displaystyle k[=]M^{-1/6}s^{-1}\)

For a first order reaction,

\(\displaystyle r=\frac{dC_{A}}{dt}=-kC_{A}\)

\(\displaystyle \frac{dC_{A}}{C_{A}}=-kdt\)

Integrating both sides gives

\(\displaystyle ln(C_{A})=-kt+C\)

Where \(\displaystyle C\) is an integration constant.

Solving for \(\displaystyle C_A\) gives

\(\displaystyle C_{A}=C_{1}e^{-kt}\)

Initially \(\displaystyle (t=0)\) there are 3 mols of \(\displaystyle A\). Using this we find that

\(\displaystyle C_{1}=3\)

\(\displaystyle C_{A}=3e^{-kt}\)

At \(\displaystyle t=500\:s\),

\(\displaystyle C_{A}=3*e^{-0.001\:s^{-1}*500\:s}=1.8196 \:mols\)

Percent conversion is calculated as follows:

\(\displaystyle X_{A}=\frac{Initial-Final}{Initial}*100 = \frac{3-1.8196}{3}*100=39\%\)

Based on the question, you can tell that the rate is first order with respect to the first reactant. So if the overall reaction is third-order, that means that the exponents must add to 3. We know that the exponent of the first reactant is 1, so that must mean that the exponent of the second reactant is 2. Thus, the concentration will be squared when you account for how it contributes to the rate, so the rate increases by a facotor of 4. 

The rate is only based on the experimental values in the table. When B doubles and A stays the same the rate doubles, making it first order with respect to compound B. When compound A doubles and B stays the same, the rate quadruples, making it second order in regards to A.

When the concentration of A stays the same but the concentration of B doubles, the rate quadruples, showing a second order rate law based on B. When the concentration of A doubles and B stays the same, the rate also doubles, showing a first order rate law based on A.

Rate laws can rarely be determined from just the reaction; they usually require experimental data. However, both of the answer choices contain the word ALWAYS, which is too extreme. Rate laws can be independent of the reactants, these rate laws are known as zero-order.

Since the rate law is provided for the reaction, we can plug in the values that are relevant to the rate of the reaction. This will allow us to determine the rate constant.

\(\displaystyle \small rate = k[Al_{2}O_{3}]^{2}[NaOH]\)

Note that HF is not included; the reaction must be zero order with respect to HF.

\(\displaystyle \small 0.005 = k[0.02]^{2}[0.02]\)

\(\displaystyle \small k = 625\)

When determining the rate law for a reaction, you need to determine the order for each reactant in the reaction. This can only be accomplished by performing an experiment where different trials show how the initial reaction rate changes based on the initial concentrations of the reactants. By keeping one initial reactant concentration constant and changing the concentration of the other reactant, we can see how the initial rate changes for each reactant. 

For reactant A, we notice that the rate quadruples when its concentration is doubled by comparing trials 2 and 3. The concentration of A increases from 0.02 to 0.04, causing the rate to change from 0.005 to 0.020. This means that the order for reactant A is 2.

\(\displaystyle \frac{rate_{2A}}{rate_A}=2^x\)

\(\displaystyle \frac{0.020}{0.005}=4=2^2\rightarrow x=2\)

For reactant B, we see that the rate does not change when the initial concentration of B is tripled by comparing trials 1 and 2. As a result, the reaction is 0 order wih respect to B.

\(\displaystyle \frac{rate_{3B}}{rate_{B}}=3^y\)

\(\displaystyle \frac{0.005}{0.005}=1=3^y\rightarrow y=0\)

Now that we have the orders for each reactant, we can write the rate law accordingly.

\(\displaystyle \small rate = k[A]^{2}\)

Since anything to the 0 power is 1, B is omitted from the rate law and can be considered to be equal to 1.