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:
For part a:
- You correctly used the rate formula, which is rate = delta [A] / delta t.
- To determine the change in molarities, you divided the moles of A by the volume of the flask (3.00 L) to get the initial and final molarities of A.
- The difference in molarities, delta M, should indeed be negative because the amount of A decreases over time.
- You calculated delta M to be -0.728 M, which is correct.
- Delta t is given as 6.85 seconds.
So far, your calculations are correct.
To calculate the rate of reaction with respect to A, you need to divide the change in molarity (-0.728 M) by the time interval (6.85 seconds). However, your negative sign placement is incorrect. The rate of the reaction with respect to A should be positive because it measures the decrease in concentration over time.
Therefore, the correct calculation for part a would be:
Rate_A = -(-0.728 M / 6.85 s)
= 0.106 M/s
For part b:
To find the rate of the reaction with respect to B, you cannot simply divide the moles of A by two. The stoichiometric coefficient of B is 1, while the stoichiometric coefficient of A is 2. This means that for every 2 moles of A reacted, it consumes 1 mole of B.
To calculate the rate of reaction with respect to B, you need to consider the change in moles of B over time. However, the problem states that B is in excess, which means it is not consumed completely. Therefore, the rate of reaction with respect to B is zero in this case.
For part c:
To find the rate of the reaction with respect to C, you need to consider the change in moles of C over time. The stoichiometric coefficient of C is 3, which means that for every 2 moles of A reacted, it produces 3 moles of C.
To calculate the rate of reaction with respect to C, you need to divide the change in moles of C by the time interval. You know that 2.737 moles of A remain after 6.85 seconds. This means that 4.925 - 2.737 = 2.188 moles of A have reacted.
Calculating the moles of C produced: 2.188 moles A x (3 moles C / 2 moles A) = 3.282 moles C.
Now, divide the moles of C by the time interval to get the rate:
Rate_C = (3.282 moles C) / (6.85 s)
= 0.478 M/s
So, to summarize:
a. The rate of reaction with respect to A is 0.106 M/s.
b. The rate of reaction with respect to B is 0 M/s (because B is in excess).
c. The rate of reaction with respect to C is 0.478 M/s.
Please note that these calculations assume that the reaction is first-order with respect to A. If the reaction has a different order, the rate expressions would be different.