|
Temp |
k |
|
0 °C |
6.6667 x 10-5 |
|
28°C |
2.0560 x 10-4 |
|
35°C |
9.6155 x 10-4 |
Homework Answers
It is not practicable to measure the intercept directly at 1/T=0, so calculate ln A from the meas...
It is not practicable to measure the intercept directly at
1/T=0, so calculate ln A from the measured slope and the
coordinates of any point on the graph. Compare your values of E and
A with those obtained by previous students, namely E = 52 kJ
mol-1 and A = ~107 dm3
mol-1 s-1
For E = 45.3953 kJ mol-1
equation: y = -5460.1x + 10.272
At 273K, 301K and 308K, I calculate the A, but it's ~3 x
10-3...
A series of experiments were conducted where the equilibrium constant (K) was determined at various temperatures...
A series of experiments were conducted where the equilibrium constant (K) was determined at various temperatures (T). The data was then graphed (shown below). The equation of the line is listed on the graph. Use the graph to determine AH° (in kJ/mol). -30 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 -30.50.003 -31 y = -6861.2x - 9.2495 -31.5 -32 -32.5 In -33 -33.5 -34 -34.5 -35 1/T in K
The rate data of the reaction was collected at different temperatures (°C) and functions of the...
The rate data of the reaction was collected at different temperatures (°C) and functions of the data plotted to determine the activation energy. 18 17 T 16.5 36.2 47.7 57.3 k 9.86 x 104 2.01 x 106 9.84 x 106 3.41 x 107 16 y = (-1.37 x 10^)x + 58.8 15 14 13 12 = 11 10 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 1. Clearly label the x and y axes (above) with the proper function. 2. Determine Ea...
*****Help me find Ea in j/moles with the slope of -5056.2!! Experiment 4 CHM 112 8....
*****Help me find Ea in j/moles with the slope of
-5056.2!!
Experiment 4 CHM 112 8. Use the following data to prepare an Arrhenius plot to determine activation energy. Follow the excel directions in the procedure section of this lab. a. Use the concentrations from mixture 1 in question 1 to calculate the rate constant, k b. Complete the rest of the table. c. in excel, plot Ink) vs 1/Temp. Use the graph to determine the slope and then the...
Run at 25.00 °C, the rate constant was found to be 2.300 X 10^-5 1/M sec....
Run at 25.00 °C, the rate constant was found to be 2.300 X 10^-5 1/M sec. Run at 30.00 °C, the rate constant was found to be 3.621 X 10^-5 1/M sec. Calculate the activation energy (Use R=8.315 J/K mol), with units of KJ/mole, and report the answer to 3 significant figures.
At 35°C, K = 2.3 x 10-5 for the reaction 2 NOCI(9) – 2 NO(g) +...
At 35°C, K = 2.3 x 10-5 for the reaction 2 NOCI(9) – 2 NO(g) + Cl (9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.4 moles of pure NOCl in a 2.0-L flask [NOCI) - t t M t [NO] - M t [Cl] - M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI] - M [NO] - M (Cl) - M c....
At 25°C the Ka of formic acid (HCO2H) is 1.77 x 10−4. Calculate the ΔG° (in...
At 25°C the Ka of formic acid (HCO2H) is 1.77 x 10−4. Calculate the ΔG° (in kJ/mol) for the ionization of formic acid in water? a. 4.63 b. 25.7 c. 1.77 x 10-4 d. 21.4 e. 3.75
At 35°C, K =2.0 x 10-8 for the reaction 2 NOCI(9) = 2 NO(g) + Cl2(9)...
At 35°C, K =2.0 x 10-8 for the reaction 2 NOCI(9) = 2 NO(g) + Cl2(9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.2 moles of pure NOCI in a 2.0-L flask [NOCI) = M [NO] = | М [Cl] = M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI) = M [NO] = M (Cl2] = 1 M c. 2.0 mole of NOCl...
22. Consider the following equilibrium: 2NOCI(8) = 210g)+Ch ) with K = 1.6 x 10-5 1.00...
22. Consider the following equilibrium: 2NOCI(8) = 210g)+Ch ) with K = 1.6 x 10-5 1.00 mole of pure NACI and 9.40 x 10-1 mole of pure Cl2 are placed in a 1.00-L container. Calculate the equilibrium concentration of Cl2(g). A) 1.6 x 10-SM B) 9.4 x 10-M C) 4.7 x 10-1 M D) 2.1 x 10-3M E) 4.1 x 10-3M
The equilibrium constant for the equation HNO3(aq) + H2O(l) = H, 0+ (aq) + NO2 (aq) is Kg = 5.6 x 10-4 M at 25.0 °C. Ca...
The equilibrium constant for the equation HNO3(aq) + H2O(l) = H, 0+ (aq) + NO2 (aq) is Kg = 5.6 x 10-4 M at 25.0 °C. Calculate the value of AGixn at 25.0 °C for nitrous acid when [NO2] = [H3O+] = [HNO2] = 1.00 M. AGixn = kJ/mol Will the acid spontaneously dissociate under these conditions? O no O yes Calculate the value of AGrxn for nitrous acid when [NO, ] = [H,O+] = 7.13 x 10- M and...
It is not practicable to measure the intercept directly at
1/T=0, so calculate ln A from the measured slope and the
coordinates of any point on the graph. Compare your values of E and
A with those obtained by previous students, namely E = 52 kJ
mol-1 and A = ~107 dm3
mol-1 s-1
For E = 45.3953 kJ mol-1
equation: y = -5460.1x + 10.272
At 273K, 301K and 308K, I calculate the A, but it's ~3 x
10-3...
A series of experiments were conducted where the equilibrium constant (K) was determined at various temperatures (T). The data was then graphed (shown below). The equation of the line is listed on the graph. Use the graph to determine AH° (in kJ/mol). -30 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 -30.50.003 -31 y = -6861.2x - 9.2495 -31.5 -32 -32.5 In -33 -33.5 -34 -34.5 -35 1/T in K
The rate data of the reaction was collected at different temperatures (°C) and functions of the data plotted to determine the activation energy. 18 17 T 16.5 36.2 47.7 57.3 k 9.86 x 104 2.01 x 106 9.84 x 106 3.41 x 107 16 y = (-1.37 x 10^)x + 58.8 15 14 13 12 = 11 10 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 1. Clearly label the x and y axes (above) with the proper function. 2. Determine Ea...
*****Help me find Ea in j/moles with the slope of
-5056.2!!
Experiment 4 CHM 112 8. Use the following data to prepare an Arrhenius plot to determine activation energy. Follow the excel directions in the procedure section of this lab. a. Use the concentrations from mixture 1 in question 1 to calculate the rate constant, k b. Complete the rest of the table. c. in excel, plot Ink) vs 1/Temp. Use the graph to determine the slope and then the...
At 35°C, K = 2.3 x 10-5 for the reaction 2 NOCI(9) – 2 NO(g) + Cl (9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.4 moles of pure NOCl in a 2.0-L flask [NOCI) - t t M t [NO] - M t [Cl] - M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI] - M [NO] - M (Cl) - M c....
At 35°C, K =2.0 x 10-8 for the reaction 2 NOCI(9) = 2 NO(g) + Cl2(9) Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.2 moles of pure NOCI in a 2.0-L flask [NOCI) = M [NO] = | М [Cl] = M b. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0-L flask [NOCI) = M [NO] = M (Cl2] = 1 M c. 2.0 mole of NOCl...
22. Consider the following equilibrium: 2NOCI(8) = 210g)+Ch ) with K = 1.6 x 10-5 1.00 mole of pure NACI and 9.40 x 10-1 mole of pure Cl2 are placed in a 1.00-L container. Calculate the equilibrium concentration of Cl2(g). A) 1.6 x 10-SM B) 9.4 x 10-M C) 4.7 x 10-1 M D) 2.1 x 10-3M E) 4.1 x 10-3M
The equilibrium constant for the equation HNO3(aq) + H2O(l) = H, 0+ (aq) + NO2 (aq) is Kg = 5.6 x 10-4 M at 25.0 °C. Calculate the value of AGixn at 25.0 °C for nitrous acid when [NO2] = [H3O+] = [HNO2] = 1.00 M. AGixn = kJ/mol Will the acid spontaneously dissociate under these conditions? O no O yes Calculate the value of AGrxn for nitrous acid when [NO, ] = [H,O+] = 7.13 x 10- M and...
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