Consider the reaction described by the equationC,H,Br, (aq) + 3 I (aq)C,H, (g) + 2 Br¯(aq) + I5 (aq)The rate law israte =k[C,H¸Br,][I¯]where k = 0.00504 M-1. s-1.What are the missing entries in the table?Run [C,H,Br,h (M). [I¯], (M)Initial rate of formation of C,H, (M-s)10.1620.0005290.162y0.00026530.1620.162MX =My =z =M-s-1

The rate of the reaction is directly proportional to the concentration of the reactants. So rate…

Consider the reaction described by the equation C,H, Br, (aq) + 31¯ (aq) –→ C,H,(g) + 2 Br¯(aq) + I, (aq) The rate law is rate = k [C,H, Br,2][I] where k = 0.00504 M-1. s- What are the missing entries in the table? Run [C,H,Br,h (M). [I¯]o (M) Initial rate of formation of C,H, (M-s) 1 0.162 0.000529 0.162 y 0.000265 3 0.162 0.162 M X = M y = z = M-s-1

Consider the reaction described by the equation C,H, Br, (aq) + 31¯ (aq) –→ C,H,(g) + 2 Br¯(aq) + I, (aq) The rate law is rate = k [C,H, Br,2][I] where k = 0.00504 M-1. s- What are the missing entries in the table? Run [C,H,Br,h (M). [I¯]o (M) Initial rate of formation of C,H, (M-s) 1 0.162 0.000529 0.162 y 0.000265 3 0.162 0.162 M X = M y = z = M-s-1

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Consider this reaction: 2H,PO, (aq)→ P,0, (aq) + 3H,0 (aq) At a certain temperature it obeys this rate law. rate = (14,5 M)[H,PO.] Suppose a vessel contains H,PO at a concentration of 1.17 M. Calculate how long it takes for the concentration of H,PO to decrease by 84.0%. You may 4 assume no other reaction is important. Round your answer to 2 significant digits.
Bromide ion is oxidized by bromate ion in acidic solution. 5Br (aq) + BrO3 (aq)+ 6H"(aq) → 3Br,(aq) + 3H,O(1) The experimentally determined rate law is Rate = k[Br ][BrO; ]H7? What is the order of reaction with respect to each reactant species? What is the overall order of the reaction? Br order: O first order O second order O third order O fourth order O fifth order BrO3 order: O first order second order O third order
Hydrogen and nitrogen(1I) oxide react according to the following equation. 2H,(g) + 2NO(g) = N;(g) + 2H,0(g) At time = t seconds, the rate of the reaction is rate = k (H,(g) [NO(g)]² (a) Explain precisely what the square brackets around nitrogen(I) oxide, [NO(g)), represent in this context. (b) Deduce the units for the rate constant k.
Using first- and second-order integrated rate laws Consider this reaction: 2N,0, (g) → 2N,0, (g) + O, (g) At a certain temperature it obeys this rate law. rate - (0.96 M 's )[N,0,] 1 Suppose a vessel contains N,0, at a concentration of 1.23 M. Calculate how olo long it takes for the concentration of N,0, to decrease by 91.%. You may Ar assume no other reaction is important. Round your answer to 2 significant digits. Ox10
The hydrolysis of the sugar sucrose to the sugars glucose and fructose can be described as follows. C12H22011 + H20 → C;H1206 + CgH1206 This reaction follows a first-order rate equation for the disappearance of sucrose: rate = k [C,,H,,0,1] (The products of the reaction, glucose, and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules, i.e. they are 22 isomers.) 1.8x10-11 Arrhenius equation at this temperature). (Enter unrounded values.) (a) In neutral solution, k at 26°C and 8.5x10-11 -1 at 37°C. Determine the activation energy (in kJ/mol), the frequency factor (in s), and the rate constant (in s) for this equation at 48°C (assuming the kinetics remain consistent with the activation energy 108 kJ/mol frequency factor rate constant X s-1 (b) When a solution of sucrose with an initial concentration of 0.130 M reaches equilibrium, the concentration of sucrose is 1.40x10 M. How long (in s) will it take the solution to reach equilibrium…
Given the following data, determine the rate constant, k (do not use units), of the reaction: A(g) + 2 B(g) → C(g) + 2 D(g) Experiment [A] (torr) [B] (torr) Rate (torr/s) 1 500 300 2.68 2 500 100 0.298 3 250 300 0.67
At a certain temperature the rate of this reaction is second order in C1₂0, with a rate constant of 4.83M¯¹ -1 S : 2C1₂05 (g) → 2Cl₂ (g) +50₂ (g) Suppose a vessel contains C1₂05 at a concentration of 1.28M. Calculate how long it takes for the concentration of Cl₂O5 to decrease to 19.0% of its initial value. You may assume no other reaction is important. Round your answer to 2 significant digits. S x10 ? X Ś
4. The rate law for the reaction H:(g) + F2(g) 2HF(g) at 298 K is rate = k [H.][F-]. %3D Predict the effect of each of the following changes on the initial rate of the reaction. * increase decrease no change Addition of H2 at constant temperature and volume Increase in volume of the reaction vessel Addition of a catalyst O O O O O O O O O
A reaction has a rate law of Rate = (1.25 M⁻¹s⁻¹)[A][B]. What is the rate of the reaction if [A] = 0.515 M and [B] = 0.280 M?
A student ran several experiments for this reaction: 5 CH3COOH(aq) + 8 MnO4 (aq) + 24 H*(aq) → 8 Mn2 (aq) + 10 CO₂(g) + 22 H₂O(l) One of the experiments gave this data: Experiment # [CH3COOH] (M) 0.269 3 [MnO4] (M) 0.395 The student determined the rate law to be: ratek [CH3COOH)² [MnO4)² [H*] Calculate k for this reaction. I [H] (M) 0.245 initial rate (M/s) 0.228
The reaction A + 2B → C proceeds according to the rate law,rate = k[A][B]with k = 8.50 x 10-3 M-1 min-1. Assume that at the start of a particular reaction, [A] = 0.270 M, [B] = 0.280 M.What is the initial rate of reaction? What is the rate when half of B has reacted? Give units! [Remember: ln(ab)= ln(a) + ln(b)]
Consider the reaction described by the equation С,Н, Br, (aq) + 31 (аq) — C,H,(g) + 2 Br¯(aq) + I, (aq) 4 The rate law is rate = k[C,H,Br, ][I¯] where k = 0.00483 M-1. s-1. What are the missing entries in the table? Run [C,H,Br, lo (M) | [I¯]o (M) Initial rate of formation of C,H, (M-s ) 4 1 0.299 0.00173 2 0.299 y 0.000864 3 0.299 0.299