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In this study of the reaction between the iodide ion and persulphate ion: S2O8-2 + 2I-...

In this study of the reaction between the iodide ion and persulphate ion:

S2O8-2 + 2I- → I2 + 2SO4-2

we have chosen to express the rate as the change in the concentration of persulphate ion, S2O8-2, per unit time.

Since persulphate ion is being consumed in the reaction, the concentration is decreasing and the rate of change is negative.

As an equation we can write:   Rate = -ΔS2O8-2 / Δt

Give two other ways that the rate of this reaction could be expressed, explain in both words and using an equation as in the example above.

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Answer #1

Solution:

The rates for the given reaction are expressed as,

S2O8^2- + 2I- = I2 + 2 SO4^2-

The rate of reactions can be expressed as,

Rate = + Δ I2 / Δt

Here, since I2 is formed in reaction, hence rate can be expressed in positive terms.

Rate = + 1/2 ΔSO4^2- / Δt

Again, since SO4^2- is formed in reaction, hence can be expressed in positive terms.

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