It may not surprise you to know that products can show up in the rate law. When the product(s) is(are) a positive order, we describe the rate law as autocatalytic. For example, under conditions of high nitric acid concentrations, this reaction: HNO, + H₂C(OH): → HNO; + HCOOH + H₂O has a rate law: rate = k [H.C(OH).] [HNO₂] Given a rate constant of 0.0255 dm3 mol-1 s-1 and initial concentrations [HNO,] = 5.0 mol dm-3, [H.C(OH).]. = 150 mmol dm-3, [HNO:]. = 8.2 mmol dm-3, how long will it take for the methanediol concentration to fall to 52 mmol dm-3?

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Chapter11: Chemical Kinetics
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It may not surprise you to know that products can show up in the rate law. When the product(s)
is(are) a positive order, we describe the rate law as autocatalytic. For example, under conditions
of high nitric acid concentrations, this reaction:
HNO, + H₂C(OH): → HNO; + HCOOH + H₂O
has a rate law: rate = k [H.C(OH).] [HNO₂]
Given a rate constant of 0.0255 dm3 mol-1 s-1 and initial concentrations [HNO,] =
5.0 mol dm-3, [H.C(OH).]. = 150 mmol dm-3, [HNO:]. = 8.2 mmol dm-3, how long will it take
for the methanediol concentration to fall to 52 mmol dm-3?
Transcribed Image Text:It may not surprise you to know that products can show up in the rate law. When the product(s) is(are) a positive order, we describe the rate law as autocatalytic. For example, under conditions of high nitric acid concentrations, this reaction: HNO, + H₂C(OH): → HNO; + HCOOH + H₂O has a rate law: rate = k [H.C(OH).] [HNO₂] Given a rate constant of 0.0255 dm3 mol-1 s-1 and initial concentrations [HNO,] = 5.0 mol dm-3, [H.C(OH).]. = 150 mmol dm-3, [HNO:]. = 8.2 mmol dm-3, how long will it take for the methanediol concentration to fall to 52 mmol dm-3?
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