. The gas phase reaction X + Y→ Z has a rate law which is experimentally observedto follow the relationship of rate = k[X]2[Y]. If the concentration of X is tripled andthe concentration of Y is doubled, the reaction rate would be increased by a factor of:a) 6 b) 9 c) 12 d) 18 e) 36 f) 48 g) 96 h) 128

Option D is correct. 18Explanation:If you have any problem let me know here in comment box for this…

Answered: . The gas phase reaction X + Y→ Z has a…

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...

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Consider the fictitious reaction and the rate law below. What is the overall order of the reaction? What happens to the rate of the reaction when the concentration of X stays the same and the concentration of Y doubles? X (g) + Y (g) → XY (g) rate = k[X][Y]
Using the information in the table, the value of the rate constant for the reaction A(g) + 3 B(g) → C(g) + 2 D(g) is
Suppose you are studying the kinetics of the reaction between nitrogen monoxide and hydrogen gas. You perform the reaction multiple times with different starting concentrations and measure the initial rate for each, resulting in this table. Experiment [NO] (M) [H2] (M) Initial Rate (M/s) 1 0.0050 0.0020 1.25 x 10–5 2 0.010 0.0020 5.00 x 10–5 3 0.0050 0.0040 2.50 x 10–5 Based on the data, choose the correct exponents to complete the rate law. rate=k[NO]x[H2]y x=? y=?
The reaction H2 (g) + 2 NO (g) + H,O (g) is studied at a certain temperature by determining the rate of reaction at reactant concentrations. The results are as follows: N,0 (g) 2. Concentration (M) Rate [NO] (i) (ii) [H] (M/s) 0.25 0.10 0.126 0.50 0.10 0.504 (iii) 0.50 0.20 1.03 The rate law is written as Rate = k [NO]* [H]°. The exponents x and y represent the order with respect to NO and H2, respectively. Calculate the value of x and y from the data provided.
A reaction was experimentally determined to follow the rate law, rate = k[X]2 where k = 0.860/M.s. If the starting concentration of X = 0.800 M, how many seconds will it take for the concentration of X to become 0.400 M? O (3) 0.465 sec O (4) 1.45 sec O (1) 0.805 sec O (2) 1.16 sec A Moving to the next question prevents changes to this answer. # T3 7 To D *3 E C 54 R F % 5 V T G 6 MacBook Pro B Y & N 7 H U N * 00 8 J 1 ( 9 K M O L P Question 12 of 37 > = است 11
The balanced equation for a given reaction is 4 X + 3 Y → P. The following data regarding initial reaction rates were collected in the lab. What is the rate law for the reaction? Trial [X] (M) [Y] (M) Rate (M/s) 1 0.020 0.060 3.40 x 10–3 2 0.020 0.120 2.72 x 10–2 3 0.080 0.060 1.36 x 10–2
A) In the proposed reaction mechanism below, the intermediate product is: A). NO(g) B). N2O2(g) C). H2(g) D). N2O(g) B) The rate expression for a particular reaction is rate = k[A][B]^2. If the initial concentration of B is increased from 0.10M to 0.30M, the initial rate will increase by which of the following factors A) 2 B) 3 C) 6 D) 9
Consider the reaction A⟶ Products. The rate law for this reaction is rate = k[A]2 where k= 1.45 ×10-2 M-1s-1 at a particular temperature. If the initial [A] = 0.300 M, what will be the concentration of A remaining after 3.00 minutes? 0.168 M 0.0221 M 0.296 M 0.287 M
just need standard error for K
1a) A(g) + B(g) C(g) The rate law for the above reaction is: -d[A]/dt = k[A][B] The rate constant is 1.13×102 L mol-1 s-1 at 346.0°C and 3.55×103 L mol-1 s-1 at 548.0°C. Use the Arrhenius equation k = Ae-Ea/RT to: Calculate Ea for this reaction. 1b) Calculate the rate constant k at 416.0°C for this reaction. 1c) Calculate the pre-exponential factor A for this reaction.
A mechanism for a fake overall chemical reaction is shown below: Step #1: X3 (g) + QX (g) → QX2 (g) + X2 (g) Step #2: QX2 (g) + X (g) → QX (g) + X2 (g) Which species would be defined as a catalyst? X QX2 X3 X2 QX

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