To determine , the data from any of the three trials could be used. Using trial 1 and the rate law, we can solve for .

Note that the units have to be adjusted, based on the order of  and .

The rate law is written as .

Compare trials 1 and 2 to see that doubling the ammonium concentration doubles the rate. The reaction is first order for ammonium: .

Compare tirals 2 and 3 to see that tripling nitrate concentration triples the rate. The reaction is first order for nitrate: .

The final rate law is .

The sum of the orders for each reactant gives you the overall order of the reaction. In this case, the reaction is first order with regard to each reactant, and is thus second order overall.

The rate law for the given reaction will be given by the formula:

The variables and need to be determined experimentally, and the value will be given by another formula.

The coefficient is a constant, and the variables in the exponent are the activation energy and temperature.

Adding an enzyme will lower the activation energy, affecting the rate constant. Changing the temperature will also affect the rate constant. Adjusting the ratio of reactants to products will alter the concentrations of the reactants. As the reaction runs, the rate declines as the reactants are consumed until it reaches equilibrium. Changing the reactant concentrations will change the reaction rate with respect to this equilibrium. Running the reaction in reverse will result in a completely different rate law since the reactants will be altered.

The rate law will have the general formula . The exponents,  and , DO NOT necessarily correlate with the reaction stoichiometry. They must be determined from the experimental data.

To determine , compare trials 1 and 3, where  is held constant.

Since  is constant and  is the same for both experiments, by plugging in the values for rate and  for trials 1 and 3, we get this equation.

.  

This reduces to , which means .  

To find , compare trials 1 and 2, where  is held constant. Following the same type of calculation as shown above would give .

The rate law is .

The reaction table in the passage indicates that the reaction rate varies in a 1-to-1 fashion as you vary the each reactant, while holding the other constant.

The rate law is written as .

Compare trials 1 and 2 to see that doubling the ammonium concentration doubles the rate. The reaction is first order for ammonium: .

Compare trials 2 and 3 to see that tripling nitrate concentration triples the rate. The reaction is first order for nitrate: .

The final rate law is .

To find the overall order of a reaction, look at the exponents for the reaction rate law. Since A has an exponent of two and B has an exponent of one, the overall reaction has an order of three. 

For a first order reaction, use the equation:

In this formula,  is the concentration at a given time,   is the initial concentration,  is the rate constant, and is the time elapsed.

If we assume the initial concentration to be , then we can use as the final concentration. Using these values and the given rate constant, we can calculate the time.

We can look at the differences between trial 1 and trial 2 to see that the reaction is first order with respect to [A], since the rate doubles when the reactant concentration doubles. Reactants are considered first order when this 1:1 relationship between concentration and rate is observed.

We can look at trials 2 and 3 to see that [C] has no effect on the reaction, making it a zeroth order reactant and simply equal to 1 in the rate law.

Trials 3 and 4 show that when [B] doubles, the rate quadruples. This means it is second order with respect to [B], increasing the rate by the square of the increase in concentration. Since we already know that reactant C does not affect the rate, it is irrelevant that its concentration is halved.

This gives us Rate = k[A][B]².

Reactant concentration, along with activation energy and temperature, is a major determinant in reaction rate, as demonstrated in Le Chatlier's principle.

This question can be approached two different ways. One way is by looking at the equation, and the other is understanding what the variables in the equation represent.  

1) The first method is understanding the equation. Because we must understand basic log and natural log functions, we know that the natural log of a fraction is a negative number. Additionally, because we know that a reaction is spontaneous when  is negative, we can conclude that a reaction will be spontaneous when Q is less than K_eq. R is a positive constant, and T cannot be negative.

2) A second approach to this question is understanding the variables in the equation. K_eq is the ratio of product to reactant at equilibrium. Q is the reaction quotient, and represents the ratio of product to reactant at a given point in time. If Q is less than K_eq, the reaction is still too far to the left, and is still reacting to reach equilibrium. Because the reaction is trying to reach equilibrium, it will continue to spontaneously react until it reaches that point.