
extra credit Helpful Stuff Gas Arrhenius Equation: k = Ae-Ea/RT Constant: R= 8.314 J/mol K Integrated...
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...
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...
help! 1 , 2 and extra credit! please answer all.
1. You study the following reaction in a series of experiments: (CH),COH(aq) + HCl(aq) à (CH),CCl(aq) +H,0(1) At 100.C you obtained the following data: Experiment (CH.COH., MHCIL. M Initial Rate, M's 100 0.100 5.0 x 10 0.100 10 x 10 300 6.200 10x10 0.300 0.100 1.5 x 10 0.00 Please do the following: a. Write the rate law for the reaction. b. Indicate the order of the reaction with respect...
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please just Question A , b and c
please this is chemistry based on kinetics
there's no other subject
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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...
Useful Information: Please show your work and write in complete sentences. R= 8.314 J/mol K 1/[A].= kt +1/[A]. In k=-E/RT+ In A In [A] =-kt + In[A]. In kı/k2 = E/R (1/T2 - 1/11) 1. The following data were collected for the reaction: 2 C102(aq) + 2 OH (g) → C103 (aq) + ClO2 (aq) + H2O(1) Experiment [C102], M [OH-], M Initial Rate, M/s 1 0.060 0.030 0.0248 2 0 .020 0.030 0.00276 0.020 0 .090 0.00828 a. Determine...
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...
The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached. The integrated rate law for a first-order reaction is: [A]=[A]0e−kt[A]=[A]0e−kt Now say we are particularly interested in the time it would take for the concentration to become one-half of its initial value. Then we could substitute [A]02[A]02 for [A][A] and rearrange the equation to: t1/2=0.693k t1/2=0.693k This equation calculates the...
14.1 Question 3 Learning Goal: To understand how to use integrated rate laws to solve for concentration. A car starts at mile marker 145 on a highway and drives at 55 mi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours? This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145. 55 mi/hr×2 hr=110 miles traveled milemarker 145−110 miles=milemarker...
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 1 ln[A]=−kt+ln[A]0ln[A]=−kt+ln[A]0 ln[A] vs. tln[A] vs. t −k 2 1[A]= kt+1[A]01[A]= kt+1[A]0 1[A] vs. t1[A] vs. t k Part A The reactant concentration in a zero-order reaction was 8.00×10−2 MM after 130 ss and 4.00×10−2 MM after 380 ss . What is...
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...