So, I go to high school in Texas and we have this AP Chem lab due online tomorrow night by midnight and I need some serious help on these questions: 

Lab Temp 0.2°C 
Experiment Trial Time in seconds Trial 1 Trial 2 Average Time
 1 1 2 256 275 268 
2 1 2 215 226 220.5
 3 1 2 83 59 71 
4 Temp here is 35 ° C 1 83 Unnecessary

Initial concentrations: Experiment [S203^2-] [S208^2-] [I-] 
1 .0022 .0044 .2222 
2 .0022 .0088 .2222 
3 .0022 .0044 .4444 
Initial rates from experiments 1,2, and 3 from [S2O3^2-] / ∆t: Experiment [Na2S2O3] (M) Avg. Elapsed Time (s) ∆ Na2S203/ ∆t (M/s) ∆ S208^2-/ ∆ t (M/s) 
1 .0022 268 8.2X10^-6 4.10*10^-6 
2 .0022 220.5 9.97X10^-6 4.98X10^-9 
3 .0022 71 3.10X10^-5 1.55X10^-5 

Questions:

1. Determine the reaction orders to S2O8^2- using experiment 1 and experiment 2. Hint: Use m = log (rate 1/rate 2) divided by log (concentration 1/ concentration 2). 

2. Determine the reaction orders to I- using experiment 1 with experiment 3. Hint: Use n = log (rate 1/rate 3) divided by log (concentration 1/ concentration 3). 

3. What is the calculated rate law for this equation/reaction? Considering most rate orders have whole integers, what is most likely the correct rate law for this reaction?

 4. Using the reaction rate calculated, calculate the rate constant for experiments 1, 2, and 3, and then the average rate constant for the three experiments. 

5. Calculate the rate constant for experiment 4 at the elevated temperature.

 6. Why are some reactions heated in laboratories?

1. To determine the reaction orders for S2O8^2-, we can use the equation m = log (rate 1/rate 2) / log (concentration 1/concentration 2).

For experiment 1, the rate is 8.2X10^-6 M/s, and for experiment 2, the rate is 9.97X10^-6 M/s. The concentrations for S2O8^2- in experiment 1 and 2 are both 0.0022 M.

Plugging these values into the equation, we have m = log (8.2X10^-6 / 9.97X10^-6) / log (0.0022 / 0.0022).

Simplifying this expression, we find m = log (0.82) / log (1) = 0.

Therefore, the reaction order for S2O8^2- using experiment 1 and experiment 2 is 0.

2. To determine the reaction orders for I-, we can use the equation n = log (rate 1/rate 3) / log (concentration 1/concentration 3).

For experiment 1, the rate is 8.2X10^-6 M/s, and for experiment 3, the rate is 3.10X10^-5 M/s. The concentrations for I- in experiment 1 and 3 are 0.2222 M and 0.4444 M, respectively.

Plugging these values into the equation, we have n = log (8.2X10^-6 / 3.10X10^-5) / log (0.2222 / 0.4444).

Simplifying this expression, we find n = log (0.26) / log (0.5) = 2.

Therefore, the reaction order for I- using experiment 1 and experiment 3 is 2.

3. The rate law for the reaction can be determined by using the reaction orders obtained from the previous questions. From experiment 1, we found that the reaction order for S2O8^2- is 0. From experiment 1 and 3, we found the reaction order for I- is 2.

Based on this information, the calculated rate law would be rate = k[S2O8^2-]^0 [I-]^2. However, since the order for S2O8^2- is 0, it means that the concentration of S2O8^2- does not affect the rate of the reaction.

Therefore, the most likely correct rate law for this reaction is rate = k [I-]^2.

4. To calculate the rate constant, we can use the initial rate from experiments 1, 2, and 3 and the corresponding concentrations of Na2S2O3 from the same experiments.

For experiments 1, 2, and 3, the initial rate is 8.2X10^-6 M/s, 9.97X10^-6 M/s, and 3.10X10^-5 M/s, respectively. The concentration of Na2S2O3 is 0.0022 M for all experiments.

Now we can use the rate equation rate = k [Na2S2O3] to solve for the rate constant (k):

k = rate / [Na2S2O3].

For experiment 1, the rate constant (k) = 8.2X10^-6 M/s / 0.0022 M = 3.7X10^-3 s^-1.
For experiment 2, k = 9.97X10^-6 M/s / 0.0022 M = 4.53X10^-3 s^-1.
For experiment 3, k = 3.10X10^-5 M/s / 0.0022 M = 14.09X10^-3 s^-1.

To find the average rate constant, we add up the individual rate constants and divide by 3:

Average rate constant = (3.7X10^-3 s^-1 + 4.53X10^-3 s^-1 + 14.09X10^-3 s^-1) / 3 ≈ 7.77X10^-3 s^-1.

Therefore, the average rate constant for the three experiments is approximately 7.77X10^-3 s^-1.

5. In order to calculate the rate constant for experiment 4 at the elevated temperature, we need additional information about the rate at this temperature. It seems that the data provided does not include the rate for experiment 4.

Without the rate, we cannot calculate the rate constant for experiment 4 at the elevated temperature.

6. Reactions are often heated in laboratories for several reasons:
- Heating can increase the rate of a reaction by providing additional energy to break chemical bonds, leading to a faster reaction.
- Some reactions require higher temperatures to reach their activation energy and proceed at a reasonable rate.
- Heating can also help in creating a uniform and controlled environment for the reaction, ensuring consistent results.
- Some reactions may be thermally activated, meaning they specifically require heat to occur.
- Heating can also enhance the solubility of reactants, allowing for better mixing and increased reaction rates.

Overall, heating in laboratories is a common technique to manipulate the rate of a reaction and control its progress.

To answer these questions, we need to understand the concepts of reaction orders and rate laws. Let's break down each question step by step:

1. Determine the reaction orders to S2O8^2- using experiment 1 and experiment 2:

To determine the reaction order, we can use the formula m = log(rate 1/rate 2) divided by log(concentration 1/concentration 2).

For experiment 1, the rate is 8.2x10^-6 M/s and the concentration of S2O8^2- is 4.10x10^-6 M.
For experiment 2, the rate is 9.97x10^-6 M/s and the concentration of S2O8^2- is 4.98x10^-9 M.

Substituting these values into the formula, we get:
m = log(8.2x10^-6 / 9.97x10^-6) / log(4.10x10^-6 / 4.98x10^-9).

Calculating this value will give you the reaction order for S2O8^2- in terms of the concentration.

2. Determine the reaction orders to I- using experiment 1 with experiment 3:

Similar to the previous question, we use the formula n = log(rate 1/rate 3) divided by log(concentration 1/concentration 3).

For experiment 1, the rate is 8.2x10^-6 M/s, and the concentration of I- is 4.10x10^-6 M.
For experiment 3, the rate is 3.10x10^-5 M/s, and the concentration of I- is 1.55x10^-5 M.

Substituting these values into the formula, we get:
n = log(8.2x10^-6 / 3.10x10^-5) / log(4.10x10^-6 / 1.55x10^-5).

Calculating this value will give you the reaction order for I- in terms of the concentration.

3. What is the calculated rate law for this equation/reaction?

The rate law shows the relationship between the rate of the reaction and the concentrations of the reactants. To determine the rate law, we need to examine how changes in concentration affect the rate.

Based on your data, you have the initial rates from the experiments and the concentrations of the reactants. By comparing the rates and the concentrations, you can determine the relationship between them and thus establish the rate law.

4. Using the reaction rate calculated, calculate the rate constant for experiments 1, 2, and 3, and then determine the average rate constant for the three experiments.

The rate constant (k) relates the rate of the reaction to the concentrations of the reactants. To calculate the rate constant, you need to use the equation rate = k[A]^x[B]^y, where [A] and [B] are the concentrations of the reactants, and x and y are the reaction orders with respect to A and B, respectively.

Using the given data, you have the rate, concentrations, and reaction orders. Substitute these values into the rate equation and solve for the rate constant (k) for each experiment. Then, calculate the average rate constant for the three experiments by taking the average of the individual rate constants.

5. Calculate the rate constant for experiment 4 at the elevated temperature.

To calculate the rate constant for experiment 4, you need the rate for experiment 4 and the concentrations for the reactants. The rate equation, as mentioned earlier, is rate = k[A]^x[B]^y. Since you only have the temperature given, you may need additional information related to the temperature dependence of the rate constant (such as activation energy) to calculate the rate constant at the elevated temperature.

6. Why are some reactions heated in laboratories?

Reactions are heated in laboratories for several reasons:

- Increase in reaction rate: Heating increases the kinetic energy of the reactant molecules, making them move faster and collide more frequently. This increased collision frequency leads to a higher reaction rate.

- Activation energy: Many reactions require a minimum energy, called the activation energy, to start. Heating provides the necessary energy to overcome the activation energy barrier and initiate the reaction.

- Shift in equilibrium: Heating can shift the equilibrium of a reversible reaction by favoring the endothermic or exothermic direction. This enables researchers to manipulate the reaction to obtain the desired products.

- Enhanced solubility: Heating can increase the solubility of reactants, allowing for more effective mixing and reaction.

- Accelerated kinetics: Certain reactions may have low reaction rates at room temperature but can proceed more rapidly at higher temperatures. Heating facilitates faster reaction kinetics.

It's important to note that not all reactions require heating. Some reactions occur at room temperature or under specific experimental conditions without the need for external heating.