R, Overview: What\'s in this rn X | (2) Keq derivation intuitio × wwN 11274: hw

Question

R, Overview: What's in this rn X | (2) Keq derivation intuitio × wwN 11274: hw 53 reedback Studio × w.php?id-672619isStudent-1 Must Visited Home WuLibrar es a Raldiff Homepage Web Assign SmartWerkVersi n 30ECampus Gri il Pandora Internet Radi. Main Menu Master ngphys xCou bol Signln blued rlabs Prirt @Cakulatr I.Periodic Tstle Ebook Question 7 of 15 (1 point) ×Incorrect X Incorrect X Incorrect X Incorrect Tutorial Problem An important reaction in the formation of photochemical smog is the reactiobt ozone and NO: The reaction is first order in NO and Og. The rate constant of the reaction is 80.0 M-s-1 at 25°C and 3.000 × 103 M Answer at 75°C. parts (a), 'b)·(C), and (d) below. (a) If this reection were to occur in a single step, would the rate law be consistent with the observed order of the reacticn for NO and 0a? yes O nc (b) What is the value of the activation energy of the reaction? 62497.84J mol O Hint Check ATSwer Vierw Salution Pragress: ExtActvity Activity Grade: 15 Polnts Possible Started: Mariday, February 13, 2017, 04:36 PM 3:43 PM Ask me anything 2/17/2017

Explanation / Answer

a)

if one step, then include only reactants, that is NO and O3

rate = k[reactants]^n

so

rate = depends on NO and O3 alone

the answr is YES

b)

Ea:

ln(k2/K1) = Ea/R*(1/T1-1/T2)

ln(80/(3*10^3)) = Ea/8.314*(1/75+273) -1/(25+273))

Ea = ln(80/(3*10^3)) * 8.314 / (1/(75+273) -1/(25+273))

Ea = 62497.77 J/mol

c)

rate when

[NO] = 3.36*10^-5 and [O3] = 5.25*10^-5

simpl, substitute K(25°C)

Rate = -k[NO][O3]

Rate =- (80)( 3.36*10^-5)(5.25*10^-5) = -1.411*10^-7 M/s

d)

find K for

T = 10°C

ln(k2/K1) = Ea/R*(1/T1-1/T2)

ln(k2/80) = 62497.77 /8.314*(1/(25+273) - 1/(273+10))

k2 = 80*exp(62497.77 /8.314*(1/(25+273) - 1/(273+10))) = 21.009

for

T = 35C

ln(k2/K1) = Ea/R*(1/T1-1/T2)

ln(k2/80) = 62497.77 /8.314*(1/(25+273) - 1/(273+35))

k2 = 80*exp(62497.77 /8.314*(1/(25+273) - 1/(273+35))) = 181.45

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