1. what is the rate law that corresponds to the data shown for the reaction A +
ID: 697741 • Letter: 1
Question
1. what is the rate law that corresponds to the data shown for the reaction A + 3B 2C? AI M 0.15 0.15 0.45 0.12 0.24 0.24 Initial Rate M/s 0.10 0.20 1.80 a) Rate kIAlIB] b) Rate -KIAPIB] c) Rate KAIBP d) Rate kAPIB e) Rate KIAPIB 2. The decomposition of hydrogen peroxide, 2H:02 2H0 + 02, is a first order reaction with a rate constant of 6.83x10s at 50 °C. If the initial concentration of hydrogen peroxide is 0.411 M, what will its concentration be after 15 minutes? a) 0.0984 M b) 0.760 M )0.222M d) 0.407 M e) 1.32 M 3. Consider the reaction B(U 3Ag, where Kc = 10.0. If 6.0 moles of A and 6.0 moles of B are introduced into a 3.00 L flask, which of the following statements is true? a) The initial reaction mixture is already at equilibrium. b) [A] increases and [B] increases to attain equilibrium. c) [A] decreases and [B] increases to attain equilibrium. d) [A] increases and [B] decreases to attain equilibrium. e) [A] = [B] at equilibrium. 4. A 0.10 M aqueous solution of which of the following salts will have the highest pH? 5. What is the pH of a 0.46 M propanoic acid (HCsHsO2) solution? For HCsHsO2, Ka 1.3x10-5 6. If you add 4.95 g NaCHsCO2 to 250 mL of a 0.150 M CHsCO2H solution, what is the pH of a) Ca(CIO)2 b) BaBr2 c)AI(CIO4)3 d) NH4 e) RbNO3 a) 0.34 b) 2.27 c) 2.61 d) 5.22 e) 11.39 the resulting buffer solution? For CHCO2H, Ka = 1.8x10-5. a) 0.824 b) 2.79 c) 4.54 d) 4.95 e) 9.05 7. What is the molar solubility of Al(OH)s in pure water? Ksp 4.6x10-33 for Al(OH)s. a) 3.6 10°M b) 4.8x109 M c) 6.3 10° M d) 8.2x10- M e) 39 10-17 M For MgFz, Ksp 6.9 x 109 Does a MgF2 precipitate form if you mix 300.0 mL of 1.1x103M MgCl2 and 500.0 mL of 1.2x10-3 M NaF? a) no precipitate forms because Q > Ksp b) no precipitate forms because Q Ksp d) a precipitate forms e) no precipitate forms because Q = Ksp because QExplanation / Answer
1)
From the given data,
Rate = k[A]^x[B]^y
Rate1/Rate2 = k[A1]^x[B1]^y/k[A2]^x[B2]^y
0.1/0.2 = (0.15)^x(0.12)^y/((0.15)^x*(0.24)^y)
y = 1 order with respect to [B]
Rate2/Rate3 = k[A2]^x[B2]^y/k[A3]^x[B3]^y
0.2/1.8 = (0.15)^x(0.12)^y/(0.45)^x(0.24)^y
x = 2 order with respect to [A]
Rate law = k[A]^2[B]^1
answer : B
2)
2 H2O2 ------> 2H2O + O2
from the first order rate equation,
K = 2.303/t*log(initial conc/Final concentration)
6.83 * 10^-4 = 2.303/(15 * 60) *log(0.411/x)
x = final concentration of H2O2 = 0.2223 M
Answer : C
3)
B <----> 3A
initially 6/3 = 2 mol/L 6/3 = 2mol/L
at equil (2 -x) 2+3x
Keq = [A]^3/[B]
10 = (2+3x)^3/(2-x)
x = 0.2059
the initial concentrations of [A] =[B] so thry at equilibrium.
Answer : E
5)
pH = 1/2 (pKa - logC)
pH = 1/2(-log(1.3*10^-5) -log(0.46))
pH = 2.612
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