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?
To solve these questions, you will need to use the provided data and formulas. Here's a step-by-step explanation of how to answer each question:
1. To determine the reaction order with respect to S2O8^2-, you can use the formula m = log (rate 1/rate 2) divided by log (concentration 1/ concentration 2).
- First, calculate the ratio of rates between experiment 1 and experiment 2: rate 1 / rate 2 = (8.2x10^-6) / (9.97x10^-6).
- Then, calculate the ratio of concentrations between experiment 1 and experiment 2: concentration 1 / concentration 2 = (0.0044) / (0.0088).
- Finally, apply the formula: m = log (rate 1/rate 2) / log (concentration 1 / concentration 2).
2. To determine the reaction order with respect to I-, use the same process as question 1, but instead use the formula n = log (rate 1/rate 3) / log (concentration 1/ concentration 3) and calculate the ratios between experiment 1 and experiment 3.
3. The rate law can be determined by analyzing the relationship between the concentration changes and the resulting rate changes. Look for patterns in the calculated rate orders for each reactant (from questions 1 and 2). If you find whole number rate orders for each reactant, you should use those as the exponents in the rate law equation.
4. To calculate the rate constant, you can use the formula k = ∆ Na2S203 / ∆t / [Na2S2O3]^m * [S2O8^2-]^n. Plug in the values from the provided data for each experiment and solve for k. Then, calculate the average rate constant by taking the average of the three experiments.
5. For experiment 4, you need to calculate the rate constant at the elevated temperature. Unfortunately, the provided data does not include the necessary information to solve this question. There should be additional data related to the effect of temperature on the rate of reaction.
6. Reactions are often heated in laboratories for various reasons, including:
- Increasing the reaction rate: Heating the reaction increases the kinetic energy of the molecules, causing them to move faster and collide more frequently. This results in a higher reaction rate.
- Activating certain reactions: Some reactions require a certain amount of energy (activation energy) to proceed. Heating the reaction provides this extra energy, allowing the reaction to occur.
- Breaking chemical bonds: Heating can break certain chemical bonds in reactant molecules, making them more reactive and facilitating the reaction.
- Enhancing solubility: In some cases, heating can increase the solubility of reactants, allowing them to mix more effectively and react faster.
- Enhancing precision: Heating can help maintain a constant temperature during a reaction, leading to more accurate and reproducible results.