chem1c-worksheet-kinetics-answer-key

linear regression

how do you divide the rate expression for Trial 2 by that for Trial 1 and simplify? How do you solve for a?

За. Bromine and methanoic acid react in aqueous solution. Br, (aq) + HCO0Н (ад) — 2Br (aq) + 2H" (аq) + Со, (g) The reaction was monitored by measuring the volume of carbon dioxide produced as time progressed. o 5 10 15 20 25 [Source: © International Baccalaureate Organization 2019] 30 20 40 60 80 100 120 140 160 Time / s [Source: © International Baccalaureate Organization 2019] Determine from the graph the rate of reaction at 20 s, in cm³ s1, showing your working. 3b. Outline, with a reason, another property that could be monitored to measure the rate of this reaction. 3c. Describe one systematic error associated with the use of the gas syringe, and how the error affects the calculated rate. 3d. Identify one error associated with the use of an accurate stopwatch. Volume of carbon dioxide / cm T m mm mo

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The following data was collected for the reaction between ethoxide and methyl iodide T(°C) Rate (1/s) T (K) 1/T (1/K) In (rate) 168 273.15 0.00366 5.12 6 354 279.15 0.00358 5.87 12 735 285.15 0.00351 6.60 18 1463 291.15 0.00343 7.29 24 3010 297.15 0.00337 8.01 30 6250 303.15 0.00330 8.74 A plot of In (rate) vs. 1/T using the above table was made and is shown below, and a linear regression line fitted to the data. Using the provided graph, estimate the activation energy (kJ/mol) of the reaction between ethoxide and methyl iodide. Enter your number using 2 decimal places without unit. Gas constant is R = 8.3144598 J. mol 1. K-1. In(k) vs 1/T 9.30 8.80 8.30 7.80 y = -9930.5x + 41.445 R² = 0.9992 2 7.30 6.80 6.30 5.80 5.30 4.80 0.00325 0.00330 0.00335 0.00340 0.00345 0.00350 0.00355 0.00360 0.00365 0.00370 1/T (1/K) Ln(k)

X2Y2 + 2Z + 2Y→ Z2 + 2Y2X Trial [X2Y2 ], (M) [Z], (M) [У, (М) Initial Rate (M/min) 1 0.25 0.11 0.33 0.025 0.11 0.33 0.2 3 0.25 0.66 0.33 0.15 4 0.25 0.11 0.66 0.025 A) What is the rate law? B) What is the rate constant? C) What is the initial rate if you have 0.061 M of X, 0.092 M of Y, and 0.043 M of Z?

I need help finding the rate law for each graph. The first graph is in order zero, the middle is in order first and the last is in order second.

CD Page view O₂N A Read aloud 1015MSC Chemistry of Biological Systems II 2020 OH TAdd text Dra

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ENZYME CATALYSIS-lab ENZYME CONCENTRATION What chemical reaction is being catalyzed in the experiment? Label the substrate(s), enzyme, and product(s).

What is m and n

You continue with the Arrhenius Plot with the slope of the curve to be approximately -12,500 K and the Arrhenius factor of 4.367 E10 hours^-1. What is the shelf life of the drug at 22 C? What is the shelf life at 4 C? Did your coworker achieve a century shelf life in either storage condition? Be sure to show your calculations. Arrhenius Plot In (k) -8 -10 -12 -14 -16 -18 0.0026 0.0028 0.0030 1/T 0.0032 0.003

The initial velocity catalyzed by Enzyme A (red) and Enzyme B (gray) for the same substrate is plotted below as a double reciprocal plot. For the following data - answer each of the fill-in-the-blank questions. -0.75 1/Vo (min/mm) 0.5+ 0.4+ 0.3 0.2- 0.1 -0.25 0.5 A 1.0 + 1/[S] (1/mM)

Rate law experiments don’t always give data in the form of a rate in molesper second. Some of them give an amount of time necessary for areaction to proceed to a given point. The faster the rate, the less timenecessary. For the following data, assume that the time recorded for eachexperiment is the time required to consume 0.10 M of A in a hypotheticalreaction  A + B   yields C Time taken (s)  [A]0 [B]0 36.8 0.20 0.40 25.0 0.20 0.60 10.0 0.50   0.60 Determine the rate law and the rate constant for the reaction

Experiment 1: Determination of the Rate, Order, and Rate Constant Table 3:  Data for Trial 1 Water displaced in mL Time (seconds) 2 mL   49 4 mL   84 6 mL   116 8 mL   145 10 mL   174 12 mL   202 14 mL   230 16 mL   253 18 mL   280 20 mL   313 22 mL   350 Table 4:  Data for Trial 2 Water displaced in mL Time (seconds) 2 mL   78 4 mL   144 6 mL   175 8 mL   205 10 mL   255 12 mL   308 14 mL   330 16 mL   409 18 mL   463 20 mL   515 22 mL   575 Table 5:  Data for Trial 3 Water displaced in mL Time (seconds) 2 mL   53 4 mL   100 6 mL   139 8 mL   178 10 mL   213 12 mL   247 14 mL   282 16 mL   319 18 mL   363 20 mL   412 22 mL   463 help me with  this part    1. Should the rate constant (k) be the same for all three trials in this experiment?  Explain your answer.…

Experiment 1: Determination of the Rate, Order, and Rate Constant Table 3:  Data for Trial 1 Water displaced in mL Time (seconds) 2 mL   49 4 mL   84 6 mL   116 8 mL   145 10 mL   174 12 mL   202 14 mL   230 16 mL   253 18 mL   280 20 mL   313 22 mL   350 Table 4:  Data for Trial 2 Water displaced in mL Time (seconds) 2 mL   78 4 mL   144 6 mL   175 8 mL   205 10 mL   255 12 mL   308 14 mL   330 16 mL   409 18 mL   463 20 mL   515 22 mL   575   Table 5:  Data for Trial 3 Water displaced in mL Time (seconds) 2 mL   53 4 mL   100 6 mL   139 8 mL   178 10 mL   213 12 mL   247 14 mL   282 16 mL   319 18 mL   363 20 mL   412 22 mL   463 Help me with  this part   Trial 1 (Table 3) Slope ______________________ Reciprocal Slope _______________   Trial 2 (Table 4)…

Experiment 1: Determination of the Rate, Order, and Rate Constant Table 3:  Data for Trial 1 Water displaced in mL Time (seconds) 2 mL   49 4 mL   84 6 mL   116 8 mL   145 10 mL   174 12 mL   202 14 mL   230 16 mL   253 18 mL   280 20 mL   313 22 mL   350 Table 4:  Data for Trial 2 Water displaced in mL Time (seconds) 2 mL   78 4 mL   144 6 mL   175 8 mL   205 10 mL   255 12 mL   308 14 mL   330 16 mL   409 18 mL   463 20 mL   515 22 mL   575 Table 5:  Data for Trial 3 Water displaced in mL Time (seconds) 2 mL   53 4 mL   100 6 mL   139 8 mL   178 10 mL   213 12 mL   247 14 mL   282 16 mL   319 18 mL   363 20 mL   412 22 mL   463 help me with  this part    Order with respect to H2O2 ___________________   Order with respect to KI ______________________

S₂O3 +31-250²/¯ +13 Trial [S₂0,2-1, (M) [1], (M) 1 2 3 0.018 0.027 0.050 0.036 0.036 0.072 Initial Rate (M/s) 2.6 x 10-6 3.9 x 10-6 1.4 x 10-5 Which answer choice number matches the rate law expression for the reaction? 1. R = k[S₂0²-] 3. R = k[S₂03 ][-] 2. R = k[S₂03-][-]³ 4. R = K[S₂0²-]²[1]

Please upload your full solution to the following problem. For the experimental isotherm data given below, determine the Freundlich and Langmuir isotherm parameters. Apply linear regression to determine the isotherm parameters. A spreadsheet can be used for this purpose. Which of the two isotherms better describes the data? Experimental data: Carbon type: F-400 Carbon size: 200 × 400 . Chemical: Trichloroethene, C₂HCl3 (131.39 g/mol) Temperature: 13 C pH: 7.5-8 Equilibrium time: 31 days ● Ⓡ B B Sample number 123456 TCE Liquid-Phase Concentration CA. μmol/L 23.6 6.67 3.26 0.322 0.169 0.114 TCE Adsorbent-Phase Concentration 9A, μmol/g 737 450 318 121 85.2 75.8 2

For an enzyme -catalyzed reaction, KM=10.0 mmol dm-3, vmax=0.250 mmol dm²³ s-'; [E]o= 2.3 x 10-6 mmol dm-3. Calculate the catalytic efficiency (=kb/KM) of the enzyme. 0.920 dm³ mol-l s-1 3.1 x 107 dm³ mol-1 s-1 1.1 x107 dm³ mol-l s-1 0.250 dm³ mol-l s-1

Derive the rate law (d[N2O5]/dt) using Steady-State Approximations   N2O5  = NO2 + NO3   (K1) NO2 + NO3 = N2O5   (K-1) NO2 + NO3 = NO2 + O2 + NO  (K2) NO + N2O5 = 3NO2   (K3)

Using the kinetic data pictured, estimate Vmax and Km for the catalysis of Substrate 1 by the following enzyme (using the correct units).

Ql- A carbohydrate (S) decomposes in the presence of an enzyme. The change of substrate concentration with initial reaction rate It's in the table below. Evaluate the value of Km &Vmax by employing the Langmuir plot. Concentration Cs Rate rp 0.0064 0.222 0.0068 0.296 0.0089 026 0.0092 0332 0.0099 0.4

1-arr In a study of the decomposition of dinitrogen pentoxide in carbon tetrachloride solution at 30 °C N2O5- →2 NO2 + ½ O2 the concentration of N,O5 was followed as a function of time. It was found that a graph of In[N2O5] versus time in minutes gave a straight line with a slope of -6.31×10* min and a y- intercept of -1.34 . Based on this plot, the reaction O order in N,0g and the rate constant for the reaction is min-1. zero first second

The active form of the enzyme alcohol dehydrogenase exists as a homotetramer (MW = 150 kDa), and has one identical active site located on each of its four subunits. You have performed a series of enzyme kinetic experiments using 0.025 mL of an enzyme sample that is 4.0 pg/mL and determined that the following Vmax = 2.442 x 10-2umoles/min What is the turnover number of this enzyme (in units of #/sec)? 76/ sec 305 / sec 152 / sec 4579/ sec 1526 / sec O O O O

The image below contains a Michaelis Menten plot for an enzyme catalysed reaction. Based on this plot, estimate the Vmax and Km. Velocity, uM/sec 40 30 20 10 0 0 ENZYME CATALYSED REACTION 20 60 Substrate concentration, mM 40 Vmax= 39 umol/sec; Km = 14 mM O Vmax = 39 umol/sec; Km = 20 mM O Vmax = 14 umol/sec; Km = 39 mM O Vmax = 20 umol/sec; Km = 14 mM 80