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A chemistry graduate student is studying the rate of this reaction: 2HI () H2 +12) He fills a reaction vessel with HI and mea

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Answer #1

From given values, Let the rate is of first order

Integrated rate law for first order is:

ln [Ao]/[A] = kt

If after substituting different values, k remains constant for this, then reaction is of first order

So, ln (0.800 / 0.502) = k (10)

k = 0.0466

When ln (0.800 / 0.315) = k(20)

k = 0.0466

ln (0.800 / 0.198) = k(30)

k = 0.0465

ln (0.800 / 0.124) = k(40)

k = 0.0466

Which is almost constant

Hence rate Law = k [HI]

and k = 0.047 min-1

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