Chapter 12 Questions

What percentage of 146C (t1/2 = 5715 years) remains in a sample estimated to be 15230 years old?

(15230/5715) = 2.66 1/2^2.66 = 0.1576 0.1576 x 100 = 15.8%

2H3PO4→P2O5+3H2O Using the information in the following table, calculate the average rate of formation of P2O5 between 10.0 and 40.0 s. Time (s) 0 10.0 20.0 30.0 40.0 50.0 [P2O5] (M) 0 2.90×10−3 5.90×10−3 7.70×10−3 8.90×10−3 9.50×10−3

(8.90x10^-3 - 2.90x10^-3) / (40-10) = 2.00*10^-4 M/s

5Br−(aq) + BrO−3(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) The average rate of consumption of Br− is 2.06×10^−4 M/s over the first two minutes. What is the average rate of formation of Br2 during the same time interval? What is the average rate of consumption of H+ during the same time interval?

2.06×10^−4 M/s * (3) / (5) =1.24×10^-4 M/s 2.06×10^−4 M/s * (6) / (5) =2.47*10^-4 M/s

NO2Cl(g) → NO2(g) + Cl(g)Cl(g) + NO2Cl(g) → NO2(g) + Cl2(g) What is the overall reaction? What is the molecularity of the first elementary step? What is the molecularity of the second elementary step? What rate law is predicted by this mechanism if the first step is rate-determining?

2NO2Cl(g) → 2NO2(g) + Cl2(g) - NO2Cl combines, NO2 combines, Cl cancels, and Cl2 remains the same First elementary step = unimolecular Second elementary step = bimolecular Rate = k[NO2Cl]

2SO2(g) + 2NO2(g) → 2SO3(g) + 2NO(g)2NO(g) + O2(g) → 2NO2(g) Write the chemical equation for the overall reaction Identify any molecule that acts as a catalyst in this reaction Identify any molecule that acts as an intermediate in this reaction

2SO2(g) + O2(g) → 2SO3(g) (2SO2 remains constant, O2 remains constant, 2SO3 remains constant, 2NO2 and 2NO cancels) Catalyst = NO2 (reactant & product side) Intermediate = NO (both on reactant side)

The activation for the following reaction is 180 kJ/mol: C2H4(g) + H2(g) → C2H6(g) If the rate constant at 202°C is 4.13 × 10-15 M-1s-1, what is the rate constant at 177°C? 3.26 × 10-16 M-1s-1 3.09 × 10-21 M-1s-1 4.22 × 10-14 M-1s-1 1.11 × 10-21 M-1s-1

3.26 × 10-16 M-1s-1

Given the following pictures at t=0min and t=10min, what is the half-life of the reaction?

5 min

Plutonium-239 has a decay constant of 2.88×10^−5year−1. What percentage of a 239Pu sample remains after 1100 years? After 2.50×10^4 years? After 1.00×10^5 years?

Decay Rate = e^-kt D= e^(-2.88x10^-5)(1100) D= 0.9688 (100) = 96.9% D= e^(-2.88x10^-5)(2.50*10^4) D= 0.4867 (100) = 48.7% D= e^(-2.88x10^-5)(1.00*10^5) D= 0.0561 (100) = 5.61%

Pt(NH3)2Cl2(aq) + H2O(l) → Pt(NH3)2(H2O)Cl+(aq) + Cl−(aq) The rate of this reaction increases by a factor of 15 on raising the temperature from 25 ∘C to 50 ∘C. What is the value of the activation energy in kJ/mol?

Ea = ln(p2/p1) * R / (1/T1 - 1/T2) Ea = ln(15) * 8.3145 / (1/298.15 - 1/323.15) Ea = 86,774 J/mol / (1000) = 87 kJ/mol

5Br−(aq) + BrO−3(aq) + 6H+(aq) ⟶ 3Br2(aq) + 3H2O(l) The reaction is first order in Br−, first order in BrO−3, and second order in H+. What is the rate law? What is the overall reaction order? How does the reaction rate change if the H+ concentration is tripled? What is the change in rate if the concentrations of both Br− and BrO−3 are halved?

Rate = k[Br-][BrO-3][H+]^2 Reaction Order = 4 (exponents added up) If H+ is tripled, The rate will increase by a factor of 9 (3^2) If Br- and BrO-3 are halved, The rate will decrease by a factor of 4 (1/2^1 * 1/2^1)

Nitrosyl bromide decomposes at 10 ∘C: 2NOBr(g)→2NO(g)+Br2(g) Time (s) 0 10 40 120 320 [NOBr] 0.0390 0.0315 0.0175 0.00784 0.00376 Use the following kinetic data to determine the order of the reaction. Determine the value of the rate constant for consumption of NOBr

Second-order k = 0.76 (1/M*s) (1/0.0390) (1/0.0315) (1/0.0175) (1/0.00784) (1/0.00376) Graph the new points Find the slope y = mx + b m = k

The rearrangement of (CH3NC) to (CH3CN) is a first-order reaction and has a rate constant of 5.11×10−5s−1 at 472 K. If the initial concentration of CH3NC is 3.40×10−2 M, What is the molarity of CH3NC after 2.20 h ? How many minutes does it take for the CH3NC concentration to drop to 2.90×10−2 M ? How many minutes does it take for 25% of the CH3NC to react?

[A]t = [A]o (e)^-kt [A]t = (3.40x10^-2) * e^(-5.11x10^-5)(7920 seconds) [A]t = 0.0226 = 2.27*10^-2 T = ln([A]t/[A]o) / -k T = ln(2.90×10−2 M/3.40×10−2 M) / (-5.11x10^-5) T = 3,112.8 seconds = 51.9 minutes T = ln([A]t/[A]o) / -k T = ln(75/100) / (-5.11x10^-5) T = 5629.8 seconds = 93.8 minutes