Question

1A. The oxidation of nitrogen monoxide by oxygen at 25 oC 2 NO + O22 NO2 is second order in NO and first order in O2. Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n... , where '1' is understood for m, n ... (don't enter 1) and concentrations taken to the zero power do not appear. Rate = In an experiment to determine the rate law, the rate constant was determined to be 9.80×103 M-2s-1. Using this value for the rate constant, the rate of the reaction when [NO] = 8.34×10-3 M and [O2] = 1.92×10-3 M would be Ms-1.

1B. The activation energy for the gas phase decomposition of t-butyl alcohol is274 kJ.

(CH₃)₃COH  (CH₃)₂C=CH₂+ H₂O
The rate constant at 732 K is1.30×10^-5 /s. The rate constant will be8.81×10^-5 /s at K.

1C. The activation energy for the gas phase decomposition of t-butyl alcohol is274 kJ.

(CH₃)₃COH  (CH₃)₂C=CH₂+ H₂O
The rate constant at 791 K is3.89×10^-4 /s. The rate constant will be /s at 825

Answer 1

1A) The expression is:

Rate = 9.80x10 ^ 3 * [NO] ^ 2 * [O2]

Is calculated:

Rate = 9.8x10 ^ 3 * (8.34x10 ^ -3) ^ 2 * 1.92x10 ^ -3 = 1.31x10 ^ -3 M / s

1B) According to the Arhenius equation you have:

k1 / k2 = exp [(- Ea / R) * (1 / T1 - 1 / T2)]

1.30x10 ^ -5 / 8.81x10 ^ -5 = exp [(- 274000 / 8.314) * (1/732) - (1 / T2)]

T2 is cleared = 764 K

1C) The Arhenius equation calculates:

3.89x10 ^ -4 / k2 = exp [(- 274000 / 8.314) * (1/791 - 1/825)

It clears k2 = 2.17x10 ^ -3

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