We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
help! 1 , 2 and extra credit! please answer all. 1. You study the following reaction...
the extra credit problem
Helpful Stuff Gas Arrhenius Equation: k = Ae-Ea/RT Constant: R= 8.314 J/mol K Integrated Rate Laws and Half-Lives: Zero order: [A] = -kt + [A]. t1/2 =[A]/2k • First order: In[A] = -kt + In[A]. t1/2 = 0.693/k • Second order: 1/[A] = kt + 1/[A] t1/2 = 1/k[A] • Pseudo first order: Use first order integrated rate law for the pseudo-first order reactant; other reactant concentrations remain constant, but are still present in overall rate...
extra credit
Helpful Stuff Gas Arrhenius Equation: k = Ae-Ea/RT Constant: R= 8.314 J/mol K Integrated Rate Laws and Half-Lives: Zero order: [A] = -kt + [A]. t1/2 =[A]/2k • First order: In[A] = -kt + In[A]. t1/2 = 0.693/k • Second order: 1/[A] = kt + 1/[A] t1/2 = 1/k[A] • Pseudo first order: Use first order integrated rate law for the pseudo-first order reactant; other reactant concentrations remain constant, but are still present in overall rate law 1....
Helpful Stuff Gas Arrhenius Equation: k = Ae-Ea/RT Constant: R= 8.314 J/mol K Integrated Rate Laws and Half-Lives: Zero order: [A] = -kt + [A]. t1/2 =[A]/2k • First order: In[A] = -kt + In[A]. t1/2 = 0.693/k • Second order: 1/[A] = kt + 1/[A] t1/2 = 1/k[A] • Pseudo first order: Use first order integrated rate law for the pseudo-first order reactant; other reactant concentrations remain constant, but are still present in overall rate law 1. You study...
Thanks in advance
please just Question A , b and c
please this is chemistry based on kinetics
there's no other subject
please if you don't know the calculation based on that
question, allow other experts to help out for the solutions
Thanks in advance
Helpful Hints: Arrhenius Equation: k =
Ae–Ea/RT Gas Constant: R = 8.314 J/mol·K
Integrated Rate Laws and Half-Lives:
• Zero order: [A] = –kt + [A]0 t1/2 = [A]0/2k
• First order: ln[A] = –kt...
could you help with this? help is much
appreciated:)
CHEMISTRY 206 Experiment 4: A KINETIC STUDY: THE REACTION OF CRYSTAL VIOLET WITH SODIUM HYDROXIDE Prelaboratory Questions This lab is the most complicated of the experiments in Chem 206. The questions in this prelaboratory exercise illustrate the calculations you will be doing with the results you will obtain, and should help you understand the reason for steps you will follow in the lab. For this reason we are not asking you...
Please show work and explain: The reaction A + B → C is studied similarly to our study of phenolphthalein fading kinetics. The corresponding rate law is Rate = k[A]m[B]m In this particular experiment, the concentration of A is sufficiently high that the pseudo-order rate law: Rate = k1[B]n can be written, where k1 = k[A]m. If the study establishes that the reaction is first order in B (n = 1) and that the pseudo rate constant k1 has the...
Homework 8-3_1 Reaction Rate Orders, Temp Effects Reaction Rates and Rate Orders The following initial rate data was collected for the reaction: HI(g) + CHşl(g) à CH.(g) +1-(9) Experiment (HI) (CHI) Initial Rate 0.015 M0.900 M 4.01 x 10M's 0.030 M0.900 M 8.04 x 10'Ms 0.030 M0.450 M 3.99 x 10 m/s 1. What is the rate law for this reaction? a. Rate =k [HI] [CHI] b. Rate = k [HI] [C,H,I] c. Rate = k [HI] (C,H,I] d. Rate...
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line, y=mx+by=mx+b. Order Integrated Rate Law Graph Slope 0 [A]=−kt+[A]0[A]=−kt+[A]0 [A] vs. t[A] vs. t −k−k 1 ln[A]=−kt+ln[A]0ln[A]=−kt+ln[A]0 ln[A] vs. tln[A] vs. t −k−k 2 1[A]= kt+1[A]01[A]= kt+1[A]0 1[A] vs. t1[A] vs. t kk A.) The reactant concentration in a zero-order reaction was 0.100 MM after 165 ss and 4.00×10−2 MM after 305 ss . What is the...
helpp
3. For the hypothetical reaction A+B → C the following rate data (with and without catalyst) were obtained: Rate, M/s Rate, M/s ALM B., Muncatalyzed) (catalyzed) 0.100 0.200 3.51 x 10 7.14 x 10 0.100 0.100 1.75 x 10 7.14 x 10 0.0500 0.200 1.76 x 10 3.57 x 10 0.0500 0.100 8.80 x 10 3.57 x 10 Based on these data: a. Determine the uncatalyzed and catalyzed rate laws for this reaction, and the corresponding rate constants. (16...
Initial rate data are listed in the table for the reaction: NH4+ (aq) + NO2 (aq) → N2 (g) + H20 (1) AWNA Experiment (NH4+1: [NO2). Initial rate (M/s) 0.24 0.10 17.2 x 10-4 0.12 0.10 13.6 x 10-4 0.15 15.4 x 10-4 0.12 0.12 4.3 x 10-4 0.12 First determine the rate law and rate constant. Under the same initial conditions as in Experiment 4, calculate (NH4+] at 103 seconds after the start of the reaction. In this experiment,...