I would really appreciate explanations on how to complete these problems, not ju

Question

I would really appreciate explanations on how to complete these problems, not just an answer. For question one, I guessed the answer. I assumed that the appearance of NO2 doubled because the coeffecient doubled, but I am not sure if that is the answer. I know number two is fourth order because you add the exponents. I am lost for 3, 4 and 5.

o 1. The decomposition of dinitrogen pentoxide is described by the chemical equation If the rate of disappearance of N20, is equal to 1.0 molmin, what is the rate of appearance of NO? A. 4.0 molmin B., 2.0 molmin C. 1.0 movmin D. 0.25 mol/min 2. The reaction shown below has the rate law: Rate klBro, JIBr][H+l'. What is the overall order ofreaction? B, second C, third D. Ar first 3. The following set of data was obtained for the reaction below. What is the rate law for the reaction? Broo' (ag) 5 Br (ag) +6 H (ag)-3 Brdag) 3 Hizo(). H (M) Rate (MW) 0.10 8.0 10 0.10 0.10 16. 10-3 o 15 o 10 24. 10-3 o 025 5.0 103 0.10 Rate 4. The first-order reaction, SotCl. SO2 +Cl, has a rate constant equal to 2.20 x 05 at 593 K. What percentage of the initial amount ofsocL will remain after 4.50 hours? 49.0% 5. The isomerization reaction, CH,NC CHcN, is first order and the rate constant is equal to 0.46 si at600 K. What is the concentration of CHNC after 0-20 minutes if the initial concentration is0.30 M? 1 x 10'' M B. 2.4 x 10 M C, 1.1 x 101 M 1.5 x 102 M

Explanation / Answer

Solved first three problems, post multiple question to get the remaining answers

Q1)

aA + bB ----- cC + dD

Rate of Reaction = -1/a d[A]/dt = -1/b d[B]/dt = 1/c d[C]/dt = 1/d d[D]/dt

-1/2 d[N2O5]/dt = 1/4 d[NO2]/dt

Rate of NO2 apperaing = 2 * rate of N2O5 disappearing = 2.0 mol/min

2)

Rate = k[BrO3-][Br-][H+]^2

Order of reaction = 1 + 1 + 2 = 4

3)

Let the rate of reaction is given by

Rate = k[BrO3-]^a[Br-]^b[H+]^c

Rate1 = 8 * 10^(-4) = k(0.10)^a(0.10)^b(0.10)^c

Rate 2 = 1.6 * 10^(-3) = k(0.20)^a(0.10)^b(0.10)^c

Rate 3 = 2.4 * 10^(-3) = k(0.20)^a(0.15)^b(0.10)^c

Rate 4 = 5 * 10^(-3) = k(0.10)^a(0.10)^b(0.25)^c

Reaction 2/Reaction 1 we get

2 = 2^a, a=1

Reaction 4/Reaction 1

6.25 = 2.5^c,c=2

Reaction3/Reaction 2

1.5 = 1.5^b, b=1

Hence the correct answer is Option C

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