The fully filled in reaction coordinate diagram is displayed below.

The arrow marked in the question represents the activation energy, which is the energy barrier that must be overcome in order for the reactants to form products.

This reaction is also exothermic because the energy of the products is lower than that of the reactants.

Start by drawing and labeling the reaction coordinate diagram.

Since we know that  is negative, the products must be lower in energy than the reactants. 

Now, the question wants you to find the activation energy if we were to reverse the reaction. In other words, we want to start at the products and end up with the reactants. From the diagram, you can see that the reverse activation energy is just the sum of the activation energy for the forward reaction and the .

In this question, we're presented with a free energy diagram for a chemical reaction occurring under standard conditions. We're asked to identify a true statement with regards to this diagram.

Right off the bat, we can see that the free energy of the products is greater than that of the reactants. Thus, this reaction has a positive change in free energy and so it is an endergonic reaction. Recall that one expression for the value of free energy change is as follows.

From this equation, we can see that the equilibrium constant is related to the standard change in free energy. Moreover, we can see that if the standard free energy change is positive, then the value of  must be less than . Thus, we know that the reaction is favored towards the reactants.

Additionally, we cannot determine whether the entropy increases or decreases based on the diagram. The reason for this is that entropy is not the only factor that determines the change in free energy of a reaction. The change in enthalpy is also a factor that must be considered.

Increasing the amount of reactant increases the rate of reaction only if the concentration of each reactant increases and if the reactants are dissolved in a solution. This is not the case for all reactions. Heating the reaction and introducing a catalyst enable the reactants to reach the activation energy needed to proceed into the reaction. Increasing the surface area between the reactants enables more of each reactant to interact with one another. 

Remember, the rate law is always determined by the rate-determining step. This is ALWAYS the "slow" step in the reaction. In this case, the slow step is step (i).

The rate for this step alone is

And because both of the original reactants are in this rate law, it is also the rate law for the overall reaction.

The correct answer is  because the differential rate law for a zero order reaction is . Therefore, the plot has a linear relationship to the concentration of hydrogen.

Because the differential rate law for a first order reaction is  the graph will have a linear relationship to .

Because the differential rate law for a second order equation is  the graph will have a linear relationship to .

In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to . When comparing experiments 1 and 3, we see that the concentration of B was doubled, and this resulted in a doubling of the rate. Therefore, the reaction is first order with respect to . Then, we plug in all of this information into the rate law, and we get the correct answer of .

In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to . When comparing experiments 1 and 3, we see that the concentration of B was doubled, and this resulted in a doubling of the rate. Therefore, the reaction is first order with respect to . Then, we plug in all of this information into the rate law, and we get the correct answer of .

Finally, we take the values from one of the rows of the experimental data. It does not matter which one, they will all result in the same answer. In this case, I chose experiment 1.

Because the values from the table all only have one significant figure, our answer can also only have one significant figure.