Question

Consider a reaction: 2HBr(g)H2(g) + Br2(g)

Express the rate of reaction with respect to each of the reactants and products.

In the first 15.0s of this reaction, the concentration of HBr dropped from 0.500M to 0.455M. Calculate the average rate of the reaction in this time interval.

Answer 1

2HBr = H2 + Br2

rate = dC/dt

Calcualte the AVERAGE rate of reaction:

Rate HBr = (0.455-0.50) /(15) = -0.003 M HBr /s

now,

for H2; Br2, the signs must be REVERSED, since they are being PRODUCED not reacted

therefore, expect positive values

due to stoichiometry, multiply by 1/2; since for every 2 mol of HBr dissapearing, we form 1 mol of each

that is, if 1 molof HBr dissapears, then 0.5 mol of H2 and Br2 appear

therefore

rates:

Rate H2 = (-0.003 M HBr /s)*(-1/2 H2/HBr) = 0.0015 M H2/s

same for Br2

Rate H2 = (-0.003 M HBr /s)*(-1/2 Br2/HBr)) = 0.0015 M Br2/s