Chemistry 2 Test 1 chem 134 exam1 homework
In balancing a half reaction, we must first balance all atoms other than H or O. Once we have done that, as in NO3- → NO we can balance H and O by adding H2O and/or H+ if the reaction is occurring in an acidic solution. What would be the coefficient of H+ when the half reaction above is balanced? A. 2 B. 4 C. 8 D. 12
Hw 4 Q 6:(Cow) 4
We know the rate law for a particular reaction A → P is rate = 4.49x10⁻³ M-1 min⁻¹ [A]² If the initial concentration of A is 2.86 M, what is the concentration of A after 102 min?
Hw 10 Q2: (sue) 2rd order BC [A]^2 1 = kt + 1 --- --- [A]t [A]o k=4.49x10-3 M-1 min-1 [A]2 t=102 [A]o=2.86 M 1 =.8076 --- [A]t so, 1/.8076 =1.24
We obtain a precipitate when we mix aqueous solutions of iron(II) nitrate and ammonium sulfide. The net ionic equation for this reaction is: ___+(aq) + __(aq) ___(s) Replace the * in the above passage. Upper and lower case letters must be used appropriately.
Hw 4 Q 4:(Cow) Fe2S2FeS
True or False. The net ionic equation for the aqueous reaction between sulfuric acid and potassium hydroxide is H+(aq) + OH-(aq) → H2O(l) A. True B. False
Hw 4 Q 5:(Cow) true
We know the rate law for a particular reaction A P is rate = 6.17x10-3 M/min If the initial concentration of A is 2.74 M, what is the concentration of A after 240 min?
Hw 10 Q4: (sue) We see that the overall order of this reaction is zero order because there is no [A] apparent in the equation [A]t = -kt + [A]0 [A]t = -(6.17x10-3 M/min)(240 min) + 2.74 M = 1.26 M
The reaction A → P has the rate constant 7.71x10⁻³ M⁻¹ min-1. We know that the plots of [A] versus time and ln[A] versus time are curved lines, while a plot of 1/[A] versus time is a straight line. If the initial concentration of A is 2.71 M, what is the concentration of A after 72.7 minutes?
Hw 10 Q3: (sue) 1/[A] is 2nd order 1 = kt + 1 --- --- [A]t [A]o k=7.71x10⁻³ M⁻¹ t=72.7 [A]o=2.71 1 =.9256 --- [A]t 1/.9256 =1.08
(7.849x10⁻⁶)(7.9x10⁻⁷) - 8.82x10⁻¹⁴
Hw 1 Q10: (dog) (WHEN MULTIPLYING x10^: just plug into calc) 6.1x10^-12
Assuming that the measurement 0.006580 kg is properly reported, the level of uncertainty is on the order of 10⁻⁶ kg. True or False?
Hw 1 Q1: (dog) Bc the digit of uncertainty is the last digit which 10^-6 in this case is the last digit. True
How many significant figures are in the following measurement (assuming it is properly reported): 0.0377 g
Hw 1 Q2: (dog) 3
How many significant figures are in the following measurement (assuming it is properly reported): 0.90020 g
Hw 1 Q3: (dog) 5
4.59g + 0.0869 g
Hw 1 Q4: (dog) 4.68 g
4.72 g- 0.0389 g
Hw 1 Q5: (dog) 4.68
A beaker containing sugar was weighed and the mass was found to be 28.6075 g. The beaker alone was previously weighed and found to have a mass of 24.72 g. What is the mass of the sugar in the beaker?
Hw 1 Q6: (dog) 3.89 g
The density of an object is the ratio of its mass to volume. A piece of metal has a mass of 192.1 g and occupies a volume of 17.3 mL. Following the general rule for rounding when multiplying or dividing numbers with significant figures, the density is equal to:
Hw 1 Q7: (dog) 11.1 g/mL
A block of wood has dimensions of 0.00660 m x 0.05909 m x 0.09210 m. Calculate the volume of the wood.
Hw 1 Q8: (dog) 0.0000359 m3
2.1x10⁷ + 6.8285x10⁹ /64.73
Hw 1 Q9: (dog) WRITE OUT THE EXPONENTS WHEN +&- 1.058x10^8
We know the rate law for a particular reaction A → P is rate = 3.77x10⁻³ M/min Calculate the time that it takes for concentration of A to drop from 2.34 M to 0.462 M.
Hw 10 Q1: (sue) 1. what eq? well, no [A] symbol so zero eq zero order [A]t = -kt + [A]o -kt= 3.77x10-3 M/min [A]o=2.34 M [A]t= 0.462 M. solving for time so no e^x =498 min
The reaction A P has the rate constant 3.93x10-3 M min-1. We know that the plots of ln[A] versus time and 1/[A] versus time are curved lines, while a plot of [A] versus time is a straight line. If the initial concentration of A is 3.24 M, what is the concentration of A after 471 minutes?
Hw 10 Q5: (sue) In this case, [A] vs t gives a linear relationship meaning that the overall order of this reaction is zero order [A]t = -kt + [A]0 [A]t = -(3.93x10-3 M/min)(471 min) + 3.24 M = 1.39 M
We know the rate law for a particular reaction A P is rate = 7.15x10-3 M-1 min-1 [A]2 Calculate the time that it takes for the concentration of A to drop from 2.58 M to 0.748 M.
Hw 10 Q6: (sue) "[A]2" indicates 2nd rate eq so plug in values (1 / 0.748 M) - (1 / 2.58 M) t = ----------------------------- 7.15x10-3 M-1 min-1 133 min
If the rate law for a reaction A → P is rate = 8.83x10⁻³ M min⁻¹ and the initial concentration of A is 0.192 M, calculate the half-life of A.
Hw 11 Q1: (hank) The overall order of this reaction is zero order; therefore, we can use the zero order intergrated rate equation or zero order half-life equation. [A]0 t1/2= ------------ 2k 0.192 M t1/2 = ---------------------- (2 * 8.83x10-3 M min-1) 10.9 min
If the rate law for a reaction A → P is rate = 4.20x10⁻³ min⁻¹ [A] and the initial concentration of A is 0.210 M, calculate the half-life of A
Hw 11 Q2: (hank) the overall order of this reaction is first order; therefore, we can use the first order intergrated rate equation or first order half-life equation .693 t1/2= ------------- 4.20x10-3 min-1 165 min
If the rate law for a reaction A → P is rate = 7.99x10⁻³ M⁻¹ min-1 [A]² and the initial concentration of A is 0.204 M, calculate the half-life of A.
Hw 11 Q3: (hank) The overall order of this reaction is second order; therefore, we can use the second order intergrated rate equation or second order half-life equation. 1 t1/2= -------- k [A]0 1 t1/2= ------------------------------ 7.99x10⁻³ M⁻¹ min-1 * 0.204 M *(SOLVE BOTTOM FIRST) 614 min
At 25oC, the half-life, t1/2, of the following reaction is 855 minutes, A → P We know the initial concentration of A is 0.174 M and that the plots of [A] versus time and ln[A] versus time are curved lines, while a plot of 1/[A] versus time is a straight line. What is the value of the rate constant, k, of this reaction?
Hw 11 Q4: (hank) 1/[A]= 2nd order 1 k= ------------ 855 min* 0.174 M (solve bottom first) 6.72x10-3 M-1 min-1
Which of the following is false? A. at higher temperatures, more collisions have sufficient energy to overcome the energy of activation B. at higher temperatures, the energy of activation is lower C. reaction rates are generally faster at higher temperatures D. at higher temperatures, molecules collide more often
Hw 12 Q1: (nine) For every reaction, there is a certain minimum energy called the activation energy that molecules must possess for collisions to effective. b
Which of the following is NOT true about a catalyst? A. it provides an alternative reaction mechanism with a lower energy of activation B. it increases the product yield C. it is not used up D. it speeds up the reaction
Hw 12 Q2: (nine) A catalyst is a substance that increases the rate of a reaction without being consumed by it. b
Consider the following fictitious mechanism: (1) R + S ---> T (2) T + U ---> W + S _________________________ R + U ---> W What is the catalyst? A. R B. S C. W D. U E. T
Hw 12 Q3: (nine) A catalyst is a substance that increases the rate of a reaction without being consumed by it. b
Consider the following mechanism: (1) O3(g) + Cl(g) ---> O2(g) + ClO(g) (2) ClO(g) + O(g)---> Cl(g) + O2(g) __________________________________ O3(g) + O(g) ---> 2O2(g) What is the intermediate? A. O3(g) B. O2(g) C. O(g) D. ClO(g) E. Cl(g)
Hw 12 Q4: (nine) An intermediate is a species produced in one step of a mechanism and consumed in a later step. d
Consider the following mechanism: (1) H2O2(aq) + I-(aq) --> H2O(l) + IO-(aq) (2) H2O2(aq) + IO-(aq) ---> H2O(l) + O2(g) + I-(aq) _______________________________________ 2H2O2(aq) ---> 2H2O(l) + O2(g) What is the catalyst? A. H2O2(aq) B. O2(g) C. H2O(l) D. IO-(aq) E. I-(aq)
Hw 12 Q5: (nine) A catalyst is a substance that increases the rate of a reaction without being consumed by it. e
Assuming that the following reactions are elementary, which reaction(s) is(are) bimolecular? I. A → B + B II. A + B → C III. A + A → B IV. A → B A. III only B. I, II, III only C. IV only D. I, III, IV only E. II and III only
Hw 13 Q1: (four) A bimolecular reaction involves a collision between two particles (atoms, molecules, or ions). E. II and III only
Assuming that the following reactions are elementary, which reaction has a first order overall rate law? I. A → B + B II. A + B → C III. A + A → B IV. A → B A. II and III only B. I, III, IV only C. I and IV only D. IV only E. I, II, III only
Hw 13 Q2: (four) (bc 1 variables on rxn side) I and IV are both unimolecular reactions. They involve one reactant molecule (A). For unimolecular reactions, the rate law is first order. For I and IV: rate = k[A]. II and III are both bimolecular reactions. They involve two reactant molecules (A and B in II, two A molecules in III). The rate laws for these are second order overall: rate = k[A][B] for II, rate = k[A]2 for III. c
Assuming that the following reactions are elementary, which reaction has a rate law that is second order overall? I. A--> B + B II. A + B -->C III. A + A -->B IV. A-> B A. II and III only B. I and IV only C. I, II, III only D. IV only E. I, III, IV only
Hw 13 Q3: (four) (bc there are two variables on reaction side, meaning second order) Since these are elementary reactions, they are based on collisions and the rate law can be written with the orders of each reactants equal to the coefficient in the balanced equation. a
Assuming that the following reactions are elementary, which reaction has a rate that is proportional to the concentration of A (all others factors being equal)? I. A --> B + B II. A + B ---> C III. A + A ---> B IV. A---> B A. I and IV only B. I, II, and IV only C. II and III only D. III only E. all four
Hw 13 Q4: (four) Looking for reactions in which A is first order. b
Which of the following types of reactions is least likely to be elementary? A. X → Y B. 2A + 3B → 2C + D C. R + S → P + Q D. 2G → H + I E. 2M + N → O + N
Hw 13 Q5: (four) The more reactants that must collide at the same time, the less likely that the reaction is elementary. b
Consider the following fictitious mechanism: 1) R + S → W + T slow 2) R + T → W fast 2R + S → 2W overall Which is the rate law for the reaction? A. rate = k[R][T] B. rate = k[W] C. rate = k[W][T] D. rate = k[R]2[S] E. rate = k[R][S]
Hw 13 Q6: (four) The overall reaction cannot proceed any faster than this slowest rate-determining step. Since the overall rate of this reaction is determined by the slow step, it seems logical that the observed rate law is Rate = k[R][S]. In a mechanism where the first elementary step is the rate-determining step (slowest step), the overall rate law is simply expressed as the elementary rate law for that slow step. This is how sometimes zero order for reactants occurrs; they were not part of slow step. e
Consider the following fictitious mechanism: 1) 2R → S (fast, reversible) 2) S + T → W slow 3) W + T → Z fast 2R + 2T → Z overall Which is the rate law for the reaction? A. rate = k[W][T] B. rate = k[S][T] C. rate = k[R]2[T]2 D. rate = k[R]2[T] E. rate = k[R]2
Hw 13 Q7: (four) The overall reaction cannot proceed any faster than this slowest rate-determining step. Since the overall rate of this reaction is determined by the slow step, it seems logical that the observed rate law is Rate = k[S][T]. In a mechanism where the first elementary step is the rate-determining step, the overall rate law is simply expressed as the elementary rate law for that slow step. However, in this problem this is not the case; the first reaction is not the rate-determining step. The rate-determining step of this mechanism contains a reaction intermediate, S, that does not appear in the overall reaction. The experimental rate law, however, can be expressed only in terms of substances that appear in the overall reaction. In this case, it is necessary to reexpress this proposed rate law in terms of substances that appear in the overall reaction eliminating the intermediate, S. Setting the forward and reverse equilibriums of the first reaction and solving for [S], we can express [S] = k1[R]2 / k-1 which leads to the final answer, rate = k[R]2[T], containing substances that appear in the overall reaction d
Cu₂O is an ionic compound and its name is copper(II) oxide. A. correct type, correct name B. correct type, incorrect name C. incorrect type, correct name D. incorrect type, incorrect name
Hw 2 Q1: (cat) Cu +2 O +2 CuO B. correct type, incorrect name
ZnCl₂ is a molecular compound and its name is zinc chloride. A. correct type, correct name B. correct type, incorrect name C. incorrect type, correct name D. incorrect type, incorrect name
Hw 2 Q2: (cat) Metal+nonmetal=Ionic The only zinc ion known to occur in compounds is Zn2+, so we do not need to indicate its charge in roman numeral. C. incorrect type, correct name
PCl₅
Hw 2 Q3: (cat) Phosphorus pentachloride
CuCl₂
Hw 2 Q4: (cat) copper(II)chloride
iron(III) sulfide
Hw 2 Q5: (cat) Fe2S3
chromium(III) oxide
Hw 2 Q6: (cat) Cr2O3
sodium carbonate
Hw 2 Q7: (cat) Na2CO3
FeSO₄
Hw 2 Q8: (cat) Iron(II)Sulfate
The molar mass of iron(III) hydroxide is A. 96.08 g/mol B. 164.1 g/mol C. 106.9 g/mol D. 102.0 g/mol E. 120.4 g/mol
Hw 3 Q1:(Bird) Fe(OH)3 106.9 g/mol
The molar mass of Ca(NO3)2 is A. 96.08 g/mol B. 164.1 g/mol C. 106.9 g/mol D. 102.0 g/mol E. 120.4 g/mol
Hw 3 Q2:(Bird) 164.1 g/mol
Calculate the mass of 15.0 moles of C₁₇H₃₆.
Hw 3 Q3:(Bird) mol->g 3606.87 g
A 248.1-mL sample of aqueous solution contains 21.4 grams of potassium carbonate. What is the molarity of the potassium carbonate in the solution?
Hw 3 Q4:(Bird) (M)olarity=(mol)/L .624 M
How many moles of sodium carbonate are in a 41.04-mL sample of 0.761 M sodium carbonate(aq) solution?
Hw 3 Q5:(Bird) (M)olarity=(mol)/L .0312 mol sodium carbonate
A 1.50-L sample of aqueous solution contains 0.950 g KNO3 and 0.977 g K2SO4. What is the molarity of the potassium ions in the solution?
Hw 3 Q6:(Bird) (M)olarity=(mol)/L 1(.950/101.11)+2(.977/174.27) ---------------------------- 1.50 .0137 M K+
Water is added to 31.6-mL sample of 0.368 M K2SO4(aq). If the final volume is 400.0 mL, what is the molarity of the resulting solution?
Hw 3 Q7:(Bird) M*V=M*V -vol is in L, molarity is in M (DON'T FORGET THIS FORMULA) .029 M
A solution is prepared by mixing 59.0 mL of 0.491 M HCl(aq) and 73.1 mL of 0.710 M HCl(aq), then adding 567.9 mL of water. Assuming the liquid volumes are additive, calculate the molarity of HCl in the resulting solution.
Hw 3 Q8:(Bird) (M)olarity=(mol)/L .590mL (.491mol/L) + 73.1 (.710 mol/L) ---------------------------------- 59.0mL+73.1mL+567.9mL 28.96 + 51.9 ----------- 700mL .115 M
State whether the following is soluble or insoluble in water: chromium(III) hydroxide A. soluble B. insoluble
Hw 4 Q 1:(Cow) Use solubility rules, rule #5. insoluble
Consider the precipitation reactions that could occur when the following pairs of aqueous solutions are mixed: Reaction 1. silver nitrate + barium chloride Reaction 2. sodium carbonate + silver sulfate The formula Ag₂SO₄ will A. appear in the net ionic equation for reaction 1 only. B. appear in the net ionic equation for reaction 2 only. C. appear in the net ionic equation for both reactions. D. not appear in the net ionic equation for either reaction because it is a spectator ion or because it is not a precipitate E. not appear in the net ionic equation for either reaction because it is not a correct formula for the any of the ions or precipitates.
Hw 4 Q 2:(Cow) D. not appear in the net ionic equation for either reaction because it is a spectator ion or because it is not a precipitate
Consider the precipitation reactions that could occur when the following pairs of aqueous solutions are mixed: Reaction 1. iron(II) nitrate + sodium carbonate Reaction 2. calcium nitrate + sodium carbonate The formula Na+ will A. appear in the net ionic equation for reaction 1 only. B. appear in the net ionic equation for reaction 2 only. C. appear in the net ionic equation for both reactions. D. not appear in the net ionic equation for either reaction because it is a spectator ion or because it is not a precipitate E. not appear in the net ionic equation for either reaction because it is not a correct formula for the any of the ions or precipitates.
Hw 4 Q 3:(Cow) D. not appear in the net ionic equation for either reaction because it is a spectator ion or because it is not a precipitate
In balancing a half reaction, we must first balance all atoms other than H or O. Once we have done that, as in MnO42 --→ MnO2 we can balance H and O by adding H2O and/or OH- if the reaction is occurring in basic solution. When H and O are balanced in the example above, what would be the coefficient of H2O? A. 1 B. 2 C. 4 D. 8
Hw 4 Q 7:(Cow) 1st: balance eq to see what side has least O, thats the side OH goes on 2nd: and water the the opposite side 3rd: balance 2
Consider a half reaction where all the atoms are already balanced, as in 2 H++ O2 → H2O2 When this half reaction is finally balanced, what would be the coefficient of e-? A. 1 B. 2 C. 3 D. 4 E. none of the above
Hw 4 Q 8:(Cow) do not include charges if not given 2
In the redox reaction: nitrate ions + copper metal ---> copper(II) ions + nitric oxide molecules; acidic medium In the balanced equation with the simplest set of whole number coefficients the sum of all the coefficients is ____ and number of electrons transferred is ____. A. 22, 6 B. 43 ,10 C. 7 ,2 D. 5, 2 E. 13, 2
Hw 4 Q 9:(Cow) if it has a charge, it would say ions 22,6
Which of the following is a unit for reaction rate? A. hr B. L mol-1 hr-1 C. M / s-1 D. mol L-1 s-1 E. min-1
Hw 5 Q1:(Bat) D. mol L-1 s-1
Which of the following is a unit for reaction rate? A. hr-1 B. L mol-1 hr-1 C. M s-1 D. M-1 min E. M / s-1
Hw 5 Q2:(Bat) C. M s-1
What is the rate of consumption of reactant A if its concentration decreases from 7.47x10⁻³ mol/L to 2.71x10⁻³ mol/L in 229.5 seconds. A. 8.82x10⁻⁸ M/s B. 2.07x10⁻⁵ M/s C. 4.64x10⁻³ M/s D. 4.44x10⁻⁵ M/s E. 6.33x10² M/s
Hw 5 Q3:(Bat) decreases so, subtract 7.47x10-3 - 2.71x10-3 and divide by 229.5= B. 2.07x10-5 M/s
The concentration of product "P" is monitored and tabulated below. What is the rate of formation of "P" between 1.02 and 4.08 minutes? Time (minutes) [P], mmol/L 0 0 1.02 0.199 2.04 0.388 3.06 0.567 4.08 0.738
Hw 5 Q4:(Bat) y/x, Time is always the x 0.738 - 0.199 -------------- = .176 4.08 - 1.02 =.176
Consider the fictitious reaction 2 A(aq) + 2 B(aq) → 2 C(aq) + 1 D(aq) If at a given instant, B is being consumed at a rate of 0.0410 M/min, then at what rate is D being produced?
Hw 5 Q5:(Bat) Set up: D -B ---- = ---- 1Δt 2Δt Replace B w/ rate: D -(.0410) ---- = ------- 1Δt 2Δt So basically -.0410/2=-.0205, but since rate is never negative its: .0205
If the rate law for the reaction of X and Y is: Rate = k [X]³ then order of reaction with respect to X is
Hw 6 Q1:(Tom) order of reaction is the exponent beside it, determined only thru experiment: 3 3
If the rate law for the reaction of X and Y is: Rate = k [X] [Y] then order of reaction with respect to Y is
Hw 6 Q2:(Tom) order of reaction is the exponent beside it, determined only thru experiment: 1, even though it isn't shown, it cant be 0 1
Suppose the rate law for a reaction is rate = k[A]²[B]² All other factors being equal, doubling the concentration of A causes the rate to A. remain the same B. double C. quadruple D. increase 8-fold
Hw 6 Q3:(Tom) If you were to double A meaning, 2A, and its already ^2 (squared) so it would be 2^2=8 (basically, just assume the variable is already 2) Quadrupled
If a reaction involves two reactants (X and Y) and the rate law is found to be rate = 3.71x10⁻⁴ [X]⁴[Y]⁰ what is the order of reaction with respect to X?
Hw 6 Q4:(Tom) Its asking for only X 4
If a reaction involves two reactants (X and Y) and the rate law is found to be rate = 3.89x10⁻⁴ [X]¹[Y]² what is the order of reaction with respect to Y?
Hw 6 Q5:(Tom) Its asking for only y 2
If a reaction involves two reactants (X and Y) and the rate law is found to be rate = 3.89x10⁻⁴[X]¹[Y]² what is the overall order of reaction?
Hw 6 Q6:(Tom) Its asking for overall, so add the exponents, for just the X, Y because that is whats being reacted 1+2=3 3
We know the rate law for a particular reaction A P is rate = 3.77x10⁻³ min⁻¹ [A] If the initial concentration of A is 2.84 M, what is the concentration of A after 235 min?
Hw 9 Q1: (sally) 1st order ln[A]t = - kt + ln[A]o "rate"= -k = 3.77x10-3min-1 "initial concentration"= ln[A]o = 2.84 M "concentration after"= t = 235 min ln[A]t = -3.77x10-3min-1(235) + ln(2.84) M =.1578 BECAUSE YOU ARE LOOKING FOR LN 2nd ln aka e^x=1.17 1.17
We know the rate law for a particular reaction A P is rate = 4.49x10⁻³ min⁻¹ [A] Calculate the time that it takes for the concentration of A to drop from 2.36 M to 0.642 M?
Hw 9 Q2: (sally) 1st order ln[A]t = - kt + ln[A]0 -k= 4.49x10-3 min-1 [A] ln[A]0= 2.36 M ln[A]t= 0.642 M no e^x =289
The reaction A → P has the rate constant 7.71x10⁻³ min⁻¹. We know that the plots of [A] versus time and 1/[A] versus time are curved lines, while a plot of ln[A] versus time is a straight line. If the initial concentration of A is 2.71 M, what is the concentration of A after 119 minutes?
Hw 9 Q3: (sally) 1st order ln[A]t = - kt + ln[A]o -k=7.71x10-3 t=119 ln[A]0=2.71 USE e^x BC LOOKING FOR LN 1.08 M
If the rate law for the decomposition of a drug, A, is rate = 4.33x10⁻³ min⁻¹ [A] calculate the time that it takes for [A] to be 91.7% decomposed.
Hw 9 Q4: (sally) 1st order ln[A]t = - kt + ln[A]o -k=4.33x10-3 min-1 [A] ln[A]o= 100 (bc 100% will eventually be dissolved) ln[A]t=(100-91.7=8.3)= 8.3 (bc 8.3 is left) time means NO e^x =575 min
Suppose the rate law for a reaction is rate = k[X]¹[Y]² and the rate of the reaction is 0.802 M/min when [X] = 0.298 M. All other factors being equal, if [X] = 0.149 M, the rate of the reaction would be A. 0.149 M/min B. 0.401 M/min C. 0.802 M/min D. 1.60 M/min
Hw7 Q1:(bob) notice .298 to .149 is halved, so half .802 and you get (or cross mult) B. 0.401 M/min
The rate of a reaction involving reactants R and S is studied using the initial rates method and the following data were obtained. Expt. [R], M [S], M Rate, M/s 1 0.147 0.249 1.97x10⁻⁴ 2 0.147 0.498 7.88x10⁻⁴ 3 0.294 0.249 7.88x10⁻⁴ What is the order of the reaction with respect to R?
Hw7 Q2:(bob) 1. Constant = exp 1 2. doubled = exp 3 =2 3.rates change by *4 4=2^? 2^2=4 Meaning the exponent is 2 2
The rate of a reaction involving reactants R and S is studied using the initial rates method and the following data were obtained. Expt. [R], M [S], M Rate, M/s 1 0.116 0.272 4.92x10⁻⁵ 2 0.116 0.816 1.48x10⁻⁴ 3 0.348 0.272 1.33x10⁻³ What is the order of the reaction with respect to S?
Hw7 Q3:(bob) Find 2 Rxn that can cancel out the most r2 = 1.48x10⁻⁴ = (0.116) (0.816) --------- -------------- r1 = 4.92x10⁻⁵ = (0.116) (0.272) cancel out .116 solve L=3 R=3 3^x=3 x=1 1
The rate of a reaction involving reactants R and S is studied using the initial rates method and the following data were obtained. Expt. [R], M [S], M Rate, M/s 1 0.103 0.273 6.07x10⁻⁶ 2 0.103 0.136 3.74x10⁻⁷ 3 0.0515 0.273 1.52x10⁻⁶ What is the order of the reaction with respect to S?
Hw7 Q4:(bob) choose rate one with a rxn that cancels out r r1= 6.07x10⁻⁶ = 0.273 --------- ----------- r2= 3.74x10⁻⁷ = 0.136 16=2^x? x=4 4
Suppose a reaction involves two reactants (X and Y) and the rate law is found to be rate = 0.136 M⁻² min-1 [Y]³ What is the rate of the reaction when [X] = 0.147 M and [Y] = 0.249 M?
Hw7 Q5:(bob) 0.136 * (.249)^3= .0021 M/min
Suppose a reaction involves two reactants (X and Y) and the rate law is found to be rate = 0.147 M⁻³ min⁻¹ [X]² [Y]² What is the rate of the reaction when [X] = 0.149 M and [Y] = 0.208 M?
Hw7 Q6:(bob) 0.147 * (0.149)^2 * (0.208)^2= 1.41x10^-4
If the rate law for the reaction of R and P is Rate = k [R] then which of the following could be a unit for the rate constant (k)? A. L mol⁻¹ s⁻¹ B. mol L⁻¹ hr⁻¹ C. yr⁻¹ D. L² mol⁻² min⁻¹
Hw8 Good Extra Q: (snake) mol/L= mol L-1 or M (mol/L)-1= L/mol or Lmol-1 or M-1 A. 2 B. 0 C. 1 D. 3 C. yr-1
If the rate law for the reaction of R and P is Rate = k [P]² then which of the following could be a unit for the rate constant (k)? A. L mol⁻¹ hr⁻¹ B. mol L⁻¹ min⁻¹ C. s⁻¹ D. L² mol⁻² s⁻¹
Hw8 Q1: (snake) mol/L= mol L-1 or M (mol/L)-1= L/mol or Lmol-1 or M-1 A. 2 B. 0 C. 1 D. 3 A. L mol⁻¹ hr⁻¹
If the rate law for the reaction of R and P is Rate = k then which of the following could be a unit for the rate constant (k)? A. L mol⁻¹ s⁻¹ B. mol L⁻¹ hr⁻¹ C. yr⁻¹ D. L² mol⁻² min⁻¹
Hw8 Q2: (snake) mol/L= mol L-1 or M (mol/L)-1= L/mol or Lmol-1 or M-1 A. 2 B. 0 C. 1 D. -3 B. mol L⁻¹ hr⁻¹
The rate of a reaction involving reactants R and S is studied using the initial rates method and the following data were obtained. Expt. [R], M [S], M Rate, M/s 1 0.272 0.272 2.91x10⁻⁴ 2 0.272 0.544 5.82x10⁻⁴ 3 0.544 0.272 2.33x10⁻³ What is the rate constant, k, of this reaction?
Hw8 Q3: (snake) constant on top, when asked for k When asked for k: determine which is rxn is constant, this one is rxn 1 r3= 2.33x10⁻³ = 0.544 0.272 ---------- ----------- r1= 2.91x10⁻⁴ 0.272 0.272 cross out terms, and solve for each side so, Left side = 8 right =2 solve 8=2^x? ....=3 *you get (.272)^3 r2= 5.82x10⁻⁴ = 0.272 0.544 ---------- ----------- r1= 2.91x10⁻⁴ 0.272 0.272 cross out terms, and solve for each side so, Left side = 2 right =2 solve 2=2^x? ....=1 *you get (.272)^1 NOW SINCE YOU'RE SOLVING FOR K: PUT CONSTANT ON TOP 2.91x10⁻⁴ -----------------= (0.272)^3 (0.272) 0.0532
Suppose a reaction involves two reactants (X and Y) and the rate law is found to be rate = k [X]² [Y]⁴ If concentrations are expressed in mol/L, a possible unit for the rate constant,k, is Mn s⁻¹ where n is equal to_______?
Hw8 Q4: (snake) 2+4=6 (its suppose to be neg) n-1=-6 -5
Suppose a reaction involves two reactants (X and Y) and the rate law is found to be 1.41x10⁻⁴ M/min = k M⁻³ min⁻¹ [X]² [Y]² What is the rate constant, k, of the reaction when [X] = 0.149 M and [Y] = 0.208 M?
Hw8 Q5: (snake) solve top and bottom separate get same x10 and just divide it 1.41x10^-4 ----------------- = 0.149^2 * 0.208^2 .1468
