Question

15 27 nterval M--1 Part c If the volume of the reaction vessel in part (b) was 1.50 L what amount of Brz (n moles) was formed during the frst 1 r m Cortana Ask me
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Answer #1

Given: The time taken = 23 seconds

Initial concentration of HBr = [HBr]0 = 0.550 M

concentration after 23s = [HBr]t = 0.457 M

The reaction is

     2HBr(g) ---> H2(g) + Br2(g)

Rate of reaction = \frac{-1}{2}\frac{\Delta HBr}{\Delta t} = \frac{\Delta Br_{2}}{\Delta t} = \frac{\Delta H_{2}}{\Delta t}

Rate of reaction = \frac{-1}{2}\frac{[HBr]_{0} -[HBr]_{t}}{\Delta t}

Therefore rate of reaction = -1/2 (0.550 - 0.457) / 23 = 0.00404 M s-1

Therefore the average rate of reaction = 0.00404 M s-1

b) The volume = 1.50 L

time = 15 seconds

Rate = d[Br2] /dt = 0.00404 M s-1

d[Br2] = 0.00404 X 15 = 0.0606 M

Moles = Molarity X volume in litres = 0.0606 M X 1.5 L = 0.0909 moles

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Consider the following reaction: 2HBr (g) rightarrow H_2 (g) + Br_2 (g) In the first 23.0...
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