Asked by Erin
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.
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.
Answers
Answered by
DrBob222
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.
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