Consider the following reaction:

2HBr(g)---> H_2(g) + Br_2(g)
B)In the first 15.0s of this reaction, the concentration of HBr dropped from 0.510M to 0.455M. Calculate the average rate of the reaction in this time interval.
C)If the volume of the reaction vessel in part (b) was 0.500L, what amount of Br_2 (in moles) was formed during the first 15.0s of the reaction?
I got the answer for part B and it was 1.8*10^-3.

mols = M x L which will allow you to calculate mols HBr present at the beginning of the reaction as well as at the 15 second mark. The difference will be the mols HBr used. Then convert that to mols Br2 (which is the same as the mols H2). Check my thinking.

To calculate the average rate of the reaction in part B, you need to use the formula:

Average rate = (change in concentration) / (change in time)

Given that the concentration of HBr dropped from 0.510M to 0.455M in the first 15.0s, you can substitute these values into the formula to find the average rate:

Average rate = (0.455M - 0.510M) / (15.0s)

Average rate = (-0.055M) / (15.0s)

Average rate = -0.0037 M/s

The average rate of the reaction in the first 15.0s is approximately -0.0037 M/s.

Now, to calculate the amount of Br2 formed during the first 15.0s of the reaction in part C, you need to use the stoichiometric coefficients of the balanced equation.

From the balanced equation: 2HBr(g) ---> H2(g) + Br2(g)

You can see that 2 moles of HBr produce 1 mole of Br2.

Based on the change in concentration of HBr in part B (0.510 M to 0.455 M), you can calculate the moles of HBr used.

Moles of HBr used = (concentration change) x (volume)
= (0.510 M - 0.455 M) x (0.500 L)
= 0.055 M x 0.500 L
= 0.0275 moles

Since 2 moles of HBr produce 1 mole of Br2, the moles of Br2 formed will be half of the moles of HBr used.

Moles of Br2 formed = (1/2) x 0.0275 moles
= 0.01375 moles

Therefore, the amount of Br2 formed during the first 15.0s of the reaction is approximately 0.01375 moles.

To find the average rate of the reaction in part (B) and the amount of Br_2 formed in part (C), you need to use stoichiometry and the concept of reaction rates.

For part (B):
1. Calculate the change in concentration of HBr:
Δ[HBr] = [HBr]final - [HBr]initial
= 0.455 M - 0.510 M
= -0.055 M (Note: The negative sign indicates a decrease in concentration.)

2. Determine the time interval:
Δt = 15.0 s (given in the question)

3. Calculate the average rate of the reaction:
Average rate = Δ[HBr] / Δt
= -0.055 M / 15.0 s
= -3.67 x 10^-3 M/s

Therefore, the average rate of the reaction in the first 15.0 seconds is -3.67 x 10^-3 M/s.

For part (C):
1. Use the balanced equation to determine the mole-to-mole ratio between HBr and Br_2:
2HBr(g) ---> H_2(g) + Br_2(g)
From the equation, we can see that 2 moles of HBr react to form 1 mole of Br_2.

2. Calculate the moles of HBr that reacted:
Moles of HBr = concentration of HBr x volume of the reaction vessel
Moles of HBr = 0.455 M x 0.500 L
= 0.228 mol

3. Use the mole-to-mole ratio to determine the moles of Br_2 formed:
Moles of Br_2 = (0.228 mol HBr) / (2 mol HBr/1 mol Br_2)
= 0.114 mol

Therefore, the amount of Br_2 formed during the first 15.0 seconds of the reaction is 0.114 moles.

It seems like your answer for part (B) is different from the one provided here. Please double-check your calculations to ensure accuracy.