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CHEM 1250 Study Guide - Quiz Guide: Ammonia, Molar Concentration, Reagent

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Ohio State University
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Chemistry
Course
CHEM 1250
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All

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Document Summary

1. (5 pts) the rate law for the decomposition of ab2 (ab2 6 ab + b2) is r = (0. 630 m!1cs!1) [ab2]2. 2nd order in ab2; k = 0. 630 m!1cs!1. Want [ab2] after 1. 00 min (60. 0 seconds) given [ab2]0 = 3. 00 m. Use the integrated rate equation for a 2nd order reaction. Need to know the order of reaction; look at the rate law above - it is 2nd order in ab2. 1/[ab2]50. 0 s = (0. 630 m!1s!1) (60. 0 s) + 1/3. 00m. 1/[ab2]50. 0 s = 38. 13 = 38. 1 m!1 [ab2]50. 0 s = 2. 622 x 10!2 m = 2. 62 x 10!2 m (3 s. f. ) 2. (3 pts) a plot of ln(r) vs. ln[a] has a slope of 1. 5 and an intercept of 2. 350. Determine the rate constant, k, and order , n, of the reaction for the rate law, r = k[a]n. show work! r = k [a]n k = rate constant n = order.

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Related Questions

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+b.

Order Integrated Rate Law Graph Slope
0 [A]=?kt+[A]0 [A] vs. t ?k
1 ln[A]=?kt+ln[A]0 ln[A] vs. t ?k
2 1[A]= kt+1[A]0 1[A] vs. t k

Part A

The reactant concentration in a zero-order reaction was 5.00×10?2M after 175 s and 4.00×10?2M after 400 s . What is the rate constant for this reaction?

Part B

What was the initial reactant concentration for the reaction described in Part A?

Part C

The reactant concentration in a first-order reaction was 6.40×10?2M after 50.0 s and 4.60×10?3M after 60.0 s . What is the rate constant for this reaction?

Part D

The reactant concentration in a second-order reaction was 0.730 M after 210 s and 6.40×10?2M after 705 s . What is the rate constant for this reaction?

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+b. Order Integrated Rate Law Graph Slope 0 [A]=−kt+[A]0 [A] vs. t −k 1 ln[A]=−kt+ln[A]0 ln[A] vs. t −k 2 1[A]= kt+1[A]0 1[A] vs. t k Part A The reactant concentration in a zero-order reaction was 0.100 M after 200 s and 4.00×10−2 M after 310 s . What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. k0th = SubmitHintsMy AnswersGive UpReview Part Part B What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. [A]0 = SubmitHintsMy AnswersGive UpReview Part Part C The reactant concentration in a first-order reaction was 6.70×10−2 M after 50.0 s and 3.20×10−3 M after 60.0 s . What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. k1st = SubmitHintsMy AnswersGive UpReview Part Part D The reactant concentration in a second-order reaction was 0.210 M after 120 s and 6.20×10−2 M after 850 s . What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

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 35

If we were to write a formula for this calculation, we might express it as follows:

milemarker=milemarker0−(speed×time)

where milemarker is the current milemarker and milemarker0 is the initial milemarker.

Similarly, the integrated rate law for a zero-order reaction is expressed as follows:

[A]=[A]0−rate×time

or

[A]=[A]0−kt

since

rate=k[A]0=k

A zero-order reaction (Figure 1) proceeds uniformly over time. In other words, the rate does not change as the reactant concentration changes. In contrast, first-order reaction rates (Figure 2) do change over time as the reactant concentration changes.

Because the rate of a first-order reaction is nonuniform, its integrated rate law is slightly more complicated than that of a zero-order reaction.

The integrated rate law for a first-order reaction is expressed as follows:

[A]=[A]0e−kt

where k is the rate constant for this reaction.

The integrated rate law for a second-order reaction is expressed as follows:

1[A]=kt+1[A0]

where k is the rate constant for this reaction.

PART A

The rate constant for a certain reaction is k = 3.50×10−3 s−1 . If the initial reactant concentration was 0.400 M, what will the concentration be after 5.00 minutes?

Express your answer with the appropriate units.

PART B

A zero-order reaction has a constant rate of 5.00×10−4M/s. If after 75.0 seconds the concentration has dropped to 2.50×10−2M, what was the initial concentration?

Express your answer with the appropriate units.