Kinetics Question: Rate of Reaction at 0.503 M

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In summary: If B had reacted with all the A in the experiment, its final concentration would have been 0.503 M. This would have resulted in a percentage change of B's concentration of 3.8%.
  • #1
davev
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Homework Statement


Consider the following reaction at 282 K.

2 A + 2 B → C + D​

where rate = rate=k[A]2. An experiment was performed for a certain number of seconds where [A]0 = 0.000303 M and 0 = 1.63 M. A plot of ln[A] vs time had a slope of -7.27. What will the rate of this reaction be when [A] = = 0.503 M?

Rate (M/s)=

The correct answer is 0.348 M/s.

Homework Equations


I'm not sure where to start here, but I tried using these equations, because I was given the slope, and the slope for ln[A] vs time is given in a first order reaction:

ln[A]t = -kt + ln[A]o
Rate = k[A]
k = -slope


The Attempt at a Solution


Basically I just tried plugging into k = 7.27 into rate=k[A]2 were [A] and are both 0.503 M.

Thank you for the help!
 
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  • #2
Look at the values of [A] and at the beginning of the experiment. From the specified stoichiometry, how much do you think the concentration of is going to change relative to its initial value during the experiment? Will this really be significant?

Chet
 
  • #3
Oh, do I have to use an ICE table?
 
  • #4
davev said:
Oh, do I have to use an ICE table?
I don't know what an ICE table is, but, whatever it is, you don't need to use it on this problem.

Chet
 
  • #5
Chestermiller said:
I don't know what an ICE table is, but, whatever it is, you don't need to use it on this problem.

Chet

ICE as in initial concentration, change in concentration, and end concentration.

How do you solve this problem?
 
  • #6
davev said:
ICE as in initial concentration, change in concentration, and end concentration.

How do you solve this problem?
The solution to this problem starts out by reconsidering the questions I asked in post #2. Let me ask in another way. If B had reacted with all the A in the experiment, what would its final concentration have been? What percentage change would this have made in the concentration of B?

Chet
 

1. What is the rate of reaction at 0.503 M?

The rate of reaction at 0.503 M refers to the speed at which a chemical reaction occurs at a concentration of 0.503 Molar. This rate is measured in terms of the change in concentration of a reactant or product over time.

2. How is the rate of reaction affected by concentration?

The rate of reaction is directly proportional to the concentration of the reactants. This means that as the concentration of the reactants increases, the rate of reaction also increases. This is because a higher concentration of reactants leads to a higher frequency of collisions, resulting in a faster reaction rate.

3. Can the rate of reaction be negative at 0.503 M?

No, the rate of reaction cannot be negative. It is always a positive value that represents the speed at which a reaction is occurring. A negative rate would indicate that the reaction is going in the opposite direction, which is not possible.

4. How does temperature affect the rate of reaction at 0.503 M?

Temperature has a significant impact on the rate of reaction at 0.503 M. As the temperature increases, the rate of reaction also increases. This is because a higher temperature leads to an increase in the kinetic energy of the molecules, resulting in more frequent and energetic collisions between reactant molecules.

5. What is the relationship between time and the rate of reaction at 0.503 M?

The rate of reaction at 0.503 M is inversely proportional to time. This means that as the reaction progresses and the concentration of reactants decreases, the rate of reaction also decreases over time. This is because there are fewer reactant molecules available to react, leading to a slower rate of reaction.

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