Question

In: Chemistry

H2+2IBr→2HBr+I2 Time (s) I2 concentration (M) 5 1.44 15 1.6 I have already figured out that...

H2+2IBr→2HBr+I2

Time (s)	I_{2 concentration (M)}
5	1.44
15	1.6

I have already figured out that the rate of formation of I2 is .016 and the rate of formation of HBr is 0.032. but now they are asking - what is the average rate of change of H2? Remember that reactant concentrations decrease over time. And I am confused how do I solve for a reactant?

Expert Solution

H₂ + 2IBr -------- 2HBr + I₂

Time (s) I₂concentration (M)

5 1.44

15 1.60

Average rate = Δ[I₂] / Δt

Average rate = 1.60 – 1.44 / 15 – 5 = 0.16 / 10

Average rate = 0.016

Time (s) I₂ concentration (M) HBr concentration (M)

5 1.44 (2 x 1.44) = 2.88

15 1.60 (2 x 1.60) = 3.20

Average rate = Δ[HBr] / Δt

Average rate = 3.20 – 2.88 / 15 – 5 = 0.32 / 10

Average rate = 0.032

Time (s) I₂concentration (M) H₂concentration (M)

5 1.44 -1.44

15 1.60 -1.60

Average rate = Δ[H₂] / Δt

Average rate = -1.60 – (-1.44) / 15 – 5 = -1.60 + 1.44 / 10 = -0.16 / 10

Average rate = -0.016

queen_honey_blossom answered 2 years ago

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