The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble
Question:
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for astraight 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 |
1) The reactant concentration in a zero-order reaction was6.00×10?2M after 150 sand3.50×10?2M after 330 s .What is the rate constant for this reaction? Express your answer with the appropriate units.
2) What was the initialreactant concentration for the reaction described in PartA? Express your answerwith the appropriate units.
3) The reactant concentration in a first-order reaction was9.20×10?2 M after50.0 s and 7.50×10?3 M after60.0 s . What is the rate constant for thisreaction? Express your answer with the appropriateunits.
4) The reactant concentration in a second-orderreaction was 0.440 M after180 s and 5.70×10?2 M after860 s . What is the rate constant for thisreaction? Express youranswer with the appropriate units. Include an asterisk to indicatea compound unit with mulitplication, for example write aNewton-meter as N*m.
Expert Answer:
To solve these problems we will use the integrated rate laws for zero first and secondorder reactions The general form of these rate laws can be written as linear equations in the form y mx b We will use the given data to solve for the rate constant and initial reactant concentration 1 For a zeroorder reaction ... View the full answer
