could you please check my work? thanks.
For the reaction between reactants A and B below, if 4.925 moles of A is placed into a flask with excess B it is found that the amount of A remaining after 6.85 seconds is 2.737 moles.
2A (g) + B (g)-----> 3 C (g)
a. find the rate of reaction with respect to A (in Ms^-1)
b. find the rate of the reaction with respect to B
c. find the reate of the reaction with respect to C
so far i have for part a:
rate = delta [A]/delta t
i calculated the molarities of before and after:
4.925mols/3.00L = 1.64 M
2.727mols/3.00L = .912 M
delta M = .728
but that's supposed to be negative because the number of mols decreases correct?
so i got -.728 M
delta t = 6.85 (0 to 6.85 seconds)
so rate a = -(.728M/6.85 seconds)
but since reactants are involved it should be negative on the outside too, so it's
=-(-.728M/6.85 s)
=.106 Ms^-1
is that correct????
to find b woul i just divide the mols of A by two????
To check your work, let's go through each part step by step:
a. Find the rate of reaction with respect to A (in Ms^-1):
You correctly started with the rate equation: rate = Δ[A]/Δt. To calculate the change in concentration of A, you found the initial molarity of A (before) and the final molarity of A (after).
Before:
Molarity of A = 4.925 moles / 3.00 L = 1.643 M
After:
Molarity of A = 2.737 moles / 3.00 L = 0.912 M
To calculate the change in concentration:
Δ[A] = (final concentration - initial concentration)
= (0.912 M - 1.643 M)
= -0.731 M (Note: It should be negative because the concentration of A decreases)
Next, you correctly determined the time interval Δt which is 6.85 seconds.
Now, we can substitute the values into the rate equation:
Rate of reaction with respect to A = Δ[A]/Δt
= (-0.731 M) / (6.85 s)
= -0.106 Ms^-1
Your calculation is correct!
b. Find the rate of reaction with respect to B:
To find the rate of reaction with respect to B, you need to consider the stoichiometry of the reaction. From the balanced equation:
2A (g) + B (g) → 3C (g)
The stoichiometric ratio between A and B is 2:1. So, for every 2 moles of A that react, 1 mole of B is consumed.
Since you found that the initial concentration of A is 1.643 M, the initial concentration of B can be calculated based on the stoichiometric ratio:
Initial concentration of B = (1.643 M)/2 = 0.822 M
However, you have excess B, so the concentration of B does not change significantly.
Therefore, the rate of reaction with respect to B is considered negligible in this case.
c. Find the rate of the reaction with respect to C:
To find the rate of the reaction with respect to C, you can use the stoichiometric ratio between A and C from the balanced equation.
The stoichiometric ratio between A and C is 2:3. So, for every 2 moles of A that react, 3 moles of C are formed.
Using the Δ[A] value obtained in part a, the change in concentration of C can be calculated:
Δ[C] = (3/2) * Δ[A]
= (3/2) * (-0.731 M)
= -1.0965 M
Now, substitute the value of Δ[C] and Δt (6.85 seconds) into the rate equation:
Rate of reaction with respect to C = Δ[C]/Δt
= (-1.0965 M) / (6.85 s)
= -0.160 Ms^-1
Your calculation for part c is correct!
Overall, your calculations for parts a, b, and c are correct, except for the assumption you made in part b that the molarities of A and B are equal. However, since the question states that B is in excess, its concentration does not change significantly. Therefore, the rate of reaction with respect to B can be considered negligible.
To check your work, let's go through each part:
a. To find the rate of reaction with respect to A (in Ms^-1), you correctly used the rate formula: rate = Δ[A]/Δt.
You calculated the initial and final molarities of A correctly, so let's determine Δ[A]:
Δ[A] = (2.737 mol - 4.925 mol) = -2.188 mol
Note that the negative sign indicates the decrease in the amount of A.
Now, let's calculate the rate:
rate = Δ[A]/Δt = (-2.188 mol)/(6.85 s) = -0.319 Ms^-1
So, the rate of reaction with respect to A is -0.319 Ms^-1.
b. To find the rate of the reaction with respect to B, you can use the stoichiometry of the balanced equation. According to the balanced equation, the molar ratio between A and B is 2:1. Therefore, if the initial moles of A are 4.925, the initial moles of B would be half of that:
Initial moles of B = (4.925 mol A) / 2 = 2.463 mol
Since B is in excess, it doesn't change during the reaction. So, the rate of the reaction with respect to B is 0 Ms^-1.
c. To find the rate of the reaction with respect to C, you need to consider the stoichiometry again. According to the balanced equation, the molar ratio between A and C is 2:3. Therefore, you can use the Δ[A] from part a to calculate Δ[C]:
Δ[C] = (3/2) * Δ[A] = (3/2) * (-2.188 mol) ≈ -3.282 mol
The negative sign indicates a decrease in the amount of C.
Now, let's calculate the rate:
rate = Δ[C]/Δt = (-3.282 mol)/(6.85 s) ≈ -0.479 Ms^-1
So, the rate of the reaction with respect to C is approximately -0.479 Ms^-1.
In summary:
a. The rate of reaction with respect to A is -0.319 Ms^-1.
b. The rate of reaction with respect to B is 0 Ms^-1.
c. The rate of reaction with respect to C is approximately -0.479 Ms^-1.
Overall, your calculations for parts a and b are correct, but please note the adjustment for part c to account for the stoichiometry of the reaction.