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question 5,6 part 1 and 2 please Question #5 Here are the relevant data to determine the activation energy of the...
Question #5 Here are the relevant data to determine the activation energy of the reaction. Rate...
Question #5 Here are the relevant data to determine the activation energy of the reaction. Rate constant (units not shown) Temperature (K) .13 293 0.11 308 Use the data above to determine the activation energy of the reaction? 83 kJ/mol Submit This value is not correct. Please check your calculations and try again. Question #6 Here are your data for trial of Lab 7 (the trial at room temperature without the catalyst): [103 initial (S203 initial Initial Rate .0023 M...
Part 3,4 and 5 please! Question #1 Rate is defined as the change in [S2052-] over...
Part 3,4 and 5 please!
Question #1 Rate is defined as the change in [S2052-] over time. Once the blue color appears, we assume [S2052-) = 0 M, so we have: Initial rate of reaction = -d[S205] = -(0-[S205 dt initial) - [S2052 Jinitial time dt Below are your data for trials 1-6 of Lab 7. Enter the initial rate for each trial. [S2052-Jinitial Time Initial rate 2.8x10-5 M/s Trial 1 .00090 M 32 s Ok Trial 2 .00090 M...
3,4,5!!! please Question #2 Here are your data for trials 1-4 of Lab 7: [103 initial...
3,4,5!!! please
Question #2 Here are your data for trials 1-4 of Lab 7: [103 initial (S203 Jinitial Initial rate Trial 1 .0023 M 1 .00090 M .000028 M/s Trial 2 .0047 M .00090 M .000069 M/S Trial 3 .0023 M .0018 M .000052 M/S Trial 4 .0047 M .0018 M .000058 M/s The order with respect to 10, should be an integer (either 0, 1, or 2). Let's determine it using two sets of data: Part 1: Use the...
part 3,4 and 5 please! Rate is defined as the change in S205 over time. Once...
part 3,4 and 5 please!
Rate is defined as the change in S205 over time. Once the blue color appears, we assume (SOS)-OM, so we have: Initial rate of reaction = "18705) (0-[S,Os initial) (5,05 initial Below are your data for trials 1-6 of Lab 7. Enter the initial rate for each trial. [S205 initial Time Initial rate 2.8x10-5 M/S Trial 1 .00090 M 32 s Ok 6.9x105 M/s Trial 2 .00090 M 13 S Ok 5.2x10-5 M/s Trial 3...
the avg k is 0.016. How do I answer part 3? Question 5 Here are your...
the avg k is 0.016. How do I answer part 3?
Question 5 Here are your data from Part C: The Effect of Temperature Room temperature Temperature of ice bath Slope of best fit line 296 K. 274 K -0.001700 Part 1: What is the numerical value of the apparent rate constant at the colder temperature, kes of the reaction? 1.700x10-3 Ok Ok Part 2: Solve for the numerical value of kc given that ke' = ke[OH-]" = k[0.30]? 5.6x10-3...
QuestiuII 1J1 Pom X " Use the following information to determine the activation energy for the...
QuestiuII 1J1 Pom X " Use the following information to determine the activation energy for the reaction shown here: A+B C +D Temperature (oC): Rate Constant (L/mol-s) 0.0323 0.148 45 98 3 7 90 O Calculate the activation energy in kJ/mol.
Temperature (°C) 4. (12) Use the following data to determine the activation energy in kJ/mol for...
Temperature (°C) 4. (12) Use the following data to determine the activation energy in kJ/mol for a first order reaction. Sketch a rough copy of the graph 300 (with labeled axes) used to determine the 320 energy. 340 Rate constant (s:') 3.2x10-11 1.0x109 3.0x108 2.4x107 67 E =
Rate Determination and Activation Energy DATA TABLE Trial Temperature (°C) Rate constant, (S-1) 0.00157 24 0.007435...
Rate Determination and Activation Energy DATA TABLE Trial Temperature (°C) Rate constant, (S-1) 0.00157 24 0.007435 2 16 3 11 0.005563 4 1. Po above, using Temperce ahd. the rate сон dik, as they axis. 2. Determine the activation energy, Ea, by plotting the natural log of k vs. the reciprocal of absolute temperature. You can calculate 1/T (first convert T to K) and in k manually, or use Excel to do it. You will also need to make a...
Part A The activation energy of a certain reaction is 34.9 kJ/mol . At 23 ∘C ,...
Part A The activation energy of a certain reaction is 34.9 kJ/mol . At 23 ∘C , the rate constant is 0.0110s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Part B Given that the initial rate constant is 0.0110s−1 at an initial temperature of 23 ∘C , what would the rate constant be at a temperature of 120. ∘C for the same reaction described in Part A?
Part A: The activation energy of a certain reaction is 44.9 kJ/mol. At 25 ∘C, the...
Part A: The activation energy of a certain reaction is 44.9 kJ/mol. At 25 ∘C, the rate constant is 0.0120s−1. At what temperature in degrees Celsius would this reaction go twice as fast? Part B: Given that the initial rate constant is 0.0120s−1 at an initial temperature of 25 ∘C, what would the rate constant be at a temperature of 120. ∘C for the same reaction described in Part A?
Question #5 Here are the relevant data to determine the activation energy of the reaction. Rate constant (units not shown) Temperature (K) .13 293 0.11 308 Use the data above to determine the activation energy of the reaction? 83 kJ/mol Submit This value is not correct. Please check your calculations and try again. Question #6 Here are your data for trial of Lab 7 (the trial at room temperature without the catalyst): [103 initial (S203 initial Initial Rate .0023 M...
Part 3,4 and 5 please!
Question #1 Rate is defined as the change in [S2052-] over time. Once the blue color appears, we assume [S2052-) = 0 M, so we have: Initial rate of reaction = -d[S205] = -(0-[S205 dt initial) - [S2052 Jinitial time dt Below are your data for trials 1-6 of Lab 7. Enter the initial rate for each trial. [S2052-Jinitial Time Initial rate 2.8x10-5 M/s Trial 1 .00090 M 32 s Ok Trial 2 .00090 M...
3,4,5!!! please
Question #2 Here are your data for trials 1-4 of Lab 7: [103 initial (S203 Jinitial Initial rate Trial 1 .0023 M 1 .00090 M .000028 M/s Trial 2 .0047 M .00090 M .000069 M/S Trial 3 .0023 M .0018 M .000052 M/S Trial 4 .0047 M .0018 M .000058 M/s The order with respect to 10, should be an integer (either 0, 1, or 2). Let's determine it using two sets of data: Part 1: Use the...
part 3,4 and 5 please!
Rate is defined as the change in S205 over time. Once the blue color appears, we assume (SOS)-OM, so we have: Initial rate of reaction = "18705) (0-[S,Os initial) (5,05 initial Below are your data for trials 1-6 of Lab 7. Enter the initial rate for each trial. [S205 initial Time Initial rate 2.8x10-5 M/S Trial 1 .00090 M 32 s Ok 6.9x105 M/s Trial 2 .00090 M 13 S Ok 5.2x10-5 M/s Trial 3...
the avg k is 0.016. How do I answer part 3?
Question 5 Here are your data from Part C: The Effect of Temperature Room temperature Temperature of ice bath Slope of best fit line 296 K. 274 K -0.001700 Part 1: What is the numerical value of the apparent rate constant at the colder temperature, kes of the reaction? 1.700x10-3 Ok Ok Part 2: Solve for the numerical value of kc given that ke' = ke[OH-]" = k[0.30]? 5.6x10-3...
QuestiuII 1J1 Pom X " Use the following information to determine the activation energy for the reaction shown here: A+B C +D Temperature (oC): Rate Constant (L/mol-s) 0.0323 0.148 45 98 3 7 90 O Calculate the activation energy in kJ/mol.
Temperature (°C) 4. (12) Use the following data to determine the activation energy in kJ/mol for a first order reaction. Sketch a rough copy of the graph 300 (with labeled axes) used to determine the 320 energy. 340 Rate constant (s:') 3.2x10-11 1.0x109 3.0x108 2.4x107 67 E =
Rate Determination and Activation Energy DATA TABLE Trial Temperature (°C) Rate constant, (S-1) 0.00157 24 0.007435 2 16 3 11 0.005563 4 1. Po above, using Temperce ahd. the rate сон dik, as they axis. 2. Determine the activation energy, Ea, by plotting the natural log of k vs. the reciprocal of absolute temperature. You can calculate 1/T (first convert T to K) and in k manually, or use Excel to do it. You will also need to make a...
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