HELP! :)

Consider the following reaction:
3A to 2B
The average rate of appearance of B is given by deltaB divided by delta t. Comparing the rate of appearance of B and the rate of disappearance of A, we get Delta B divided by Delta t= ___* (-delta A divided by delta t)

Thanks!

Sorry got the questions mixed, i know this one because there was a table provided. But how do i find...


Consider the following reaction:
A to 2C
The average rate of appearance of C is given by delta C divided by delta t. Comparing the rate of appearance of C and the rate of disappearance of A, we get delta c / t = ____ * (delta A/ delta t).

thanks again!

The first question is 3/2?

To find the expression for the average rate of appearance of B in terms of the rate of disappearance of A, we need to use the stoichiometric coefficients of the balanced equation.

Given the reaction:
3A → 2B

The stoichiometric coefficients represent the number of moles of each substance involved in the reaction. In this case, it tells us that 3 moles of A react to form 2 moles of B.

The rate of disappearance of A can be expressed as (delta A) / (delta t), representing the change in the concentration of A over a specific time interval.

The rate of appearance of B can be expressed as (delta B) / (delta t), representing the change in the concentration of B over the same time interval.

Comparing the rate of appearance of B and the rate of disappearance of A, we can write:

(delta B) / (delta t) = (2 / 3) * (-(delta A) / (delta t))

Therefore, the expression for the average rate of appearance of B in terms of the rate of disappearance of A is:

(delta B) / (delta t) = -(2 / 3) * (delta A) / (delta t)

So, the coefficient multiplying (-delta A) / (delta t) is -2/3.

I hope this explanation helps! Let me know if you have any further questions.

To compare the rate of appearance of B and the rate of disappearance of A in the given reaction, we need to first understand the stoichiometry of the reaction. The stoichiometry tells us the molar ratio or the ratio of molecules participating in the reaction.

In this case, the reaction tells us that 3A molecules react to form 2B molecules. This means that for every 3 moles of A that disappear, 2 moles of B will appear.

Now let's consider the rate of appearance of B and the rate of disappearance of A over a specific time interval, represented by Δt. The rate of appearance of B, given by ΔB/Δt, is the change in the concentration of B divided by the change in time.

Similarly, the rate of disappearance of A, given by -ΔA/Δt, is the change in the concentration of A (with a negative sign to indicate disappearance) divided by the change in time.

Since the stoichiometry of the reaction tells us that for every 3 moles of A that disappear, 2 moles of B appear, we can use this ratio to compare their rates. The ratio will be:

(ΔB/Δt) = (2/3) * (-ΔA/Δt)

So, to get the value to fill in the blank, multiply the negative rate of disappearance of A (-ΔA/Δt) by 2/3.

I hope this explanation helps! Let me know if you have any more questions.