Find the integral rate equation of 1st order reaction

In summary, the question asks for the integral rate equation for a reaction that is 1st order with respect to A and 0 order with respect to B, starting with initial concentrations of [A0] and [B0]. The attempt at a solution provided an equation of ln[A0]/[A0]-v1x=kt, but the given answer was ln[A0]/[A0]-v1x=v1kt. The poster also requested clarification on the use of v1 and x in the equation.
  • #1
avistein
48
1

Homework Statement



The reaction v1A+v2B--->Products is 1st order wrt A and 0 order wrt B.If the reaction is started with [A0] and [B0] find the integral rate equation.

The Attempt at a Solution



I have got ln[A0]/[A0]-v1x=kt but answer is given as ln[A0]/[A0]-v1x=v1kt.Which one is right?
 
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  • #2
Please provide us with your work.
 
  • #3
ln[A0]/[A]=kt is the integrated rate equation for 1st order reaction.where A0 is intitial conc at time=0 and A is conc left after time =t .Here v1x has been used and [A0]-v1x is left.so that becomes ln[A0]/[A0]-v1x=kt?
 
  • #4
I'm sorry but I don't understand what you are doing. I don't understand what v1 is or what x is. You need to show your work carefully, not reiterate what you wrote in the first post.
 
  • #5


The correct answer is ln[A0]/[A0]-v1x=v1kt. This can be derived from the integrated rate law for a first-order reaction, which is ln[A]t = -kt + ln[A]0, where [A]t is the concentration of A at time t, k is the rate constant, and [A]0 is the initial concentration of A. In the given reaction, since the reaction is first-order with respect to A, the concentration of A decreases exponentially over time. Therefore, the integrated rate law can be rewritten as ln[A0]/[A]t = kt, where [A0] is the initial concentration of A. This equation shows the ratio of the initial concentration of A to the concentration of A at any given time, which is what the given answer, ln[A0]/[A0]-v1x=v1kt, represents. So, the given answer is correct and your attempt at a solution is incorrect.
 

FAQ: Find the integral rate equation of 1st order reaction

What is a 1st order reaction?

A 1st order reaction is a chemical reaction in which the rate of reaction is directly proportional to the concentration of only one reactant. This means that as the concentration of the reactant decreases, the rate of reaction also decreases.

How do you determine the integral rate equation for a 1st order reaction?

The integral rate equation for a 1st order reaction is ln[A] = -kt + ln[A]0, where [A] is the concentration of the reactant at a given time, k is the rate constant, t is time, and [A]0 is the initial concentration of the reactant. This equation can be derived using calculus and the integrated rate law for a 1st order reaction.

What is the significance of the rate constant in the integral rate equation for a 1st order reaction?

The rate constant, k, is a measure of the reaction rate and is specific to a particular reaction at a given temperature. It is affected by factors such as temperature, presence of a catalyst, and the nature of the reactants. A higher value of k indicates a faster reaction rate, while a lower value indicates a slower reaction rate.

Can the integral rate equation for a 1st order reaction be used to determine the concentration of a reactant at a specific time?

Yes, the integral rate equation can be rearranged to solve for the concentration of the reactant at a specific time, as long as the values of k, [A]0, and t are known. This can be useful in determining the progress of a reaction over time or predicting the concentration of a reactant at a future time.

Are there any limitations to the use of the integral rate equation for a 1st order reaction?

Yes, the integral rate equation assumes that the reaction is taking place at a constant temperature and that the rate of reaction is only dependent on the concentration of one reactant. If these conditions are not met, the integral rate equation may not accurately predict the rate of reaction. Additionally, the reaction must truly be a 1st order reaction for the equation to be valid.

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