CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM
7th Edition
ISBN: 9781319420994
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 7, Problem 7B.2AST
Interpretation Introduction
Interpretation:
The data given for the decomposition of
Concept Introduction:
According to the integrated rate law for the first order reaction, the concentration of reactant is the exponential function of time. The two equations that represent the integrated rate law for the first order kinetics is shown below.
The plot of natural logarithm of concentration of reactant against time is the straight line and the slope of the graph is
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25) Consider the following set of reactions. Assume that A, B, C and D are atoms.
(1) A + A → B
(2) А-В + С-D А-С-D-B
(3) A + B-C → A-B-C
Order these reactions in terms of increasing significance of the orientation probability (that is
the A in the Arrhenius equation).
А. (1)
(2) < (3)
В. (3) < (2) <(1)
C. (1) < (3) <(2)
D. (1) = (2) = (3)
E. (1) < (2) = (3)
(a) A reaction 2 A →P has a second-order rate law with k = 3.50 × 10−4 dm3 mol−1s−1. Calculate the time required for the concentration of A to change from 0.260 mol dm−3 to 0.011 mol dm−3
(b) For the ideal gas reaction A + B ↔ 2C + 2D, it is given that ∆Gθ500= 1250 cal mol-1. If 1.0 mol of A and 1.0 mol of B are placed in a vessel at 500 K and P is held fixed at 1200 torr, calculate the equilibrium amount of A, B, C, and D.
The reaction A− + H+ → P has a rate constant given by the empirical expression kr = Ae−(5925 K)/T with A = 6.92 × 1012 dm3 mol−1 s−1. Evaluate the enthalpy and entropy of activation at 25 °C.
Chapter 7 Solutions
CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM
Ch. 7 - Prob. 7A.1ASTCh. 7 - Prob. 7A.1BSTCh. 7 - Prob. 7A.2ASTCh. 7 - Prob. 7A.2BSTCh. 7 - Prob. 7A.3ASTCh. 7 - Prob. 7A.3BSTCh. 7 - Prob. 7A.4ASTCh. 7 - Prob. 7A.4BSTCh. 7 - Prob. 7A.1ECh. 7 - Prob. 7A.2E
Ch. 7 - Prob. 7A.3ECh. 7 - Prob. 7A.4ECh. 7 - Prob. 7A.7ECh. 7 - Prob. 7A.8ECh. 7 - Prob. 7A.9ECh. 7 - Prob. 7A.10ECh. 7 - Prob. 7A.11ECh. 7 - Prob. 7A.12ECh. 7 - Prob. 7A.13ECh. 7 - Prob. 7A.14ECh. 7 - Prob. 7A.15ECh. 7 - Prob. 7A.16ECh. 7 - Prob. 7A.17ECh. 7 - Prob. 7A.18ECh. 7 - Prob. 7B.1ASTCh. 7 - Prob. 7B.1BSTCh. 7 - Prob. 7B.2ASTCh. 7 - Prob. 7B.2BSTCh. 7 - Prob. 7B.3ASTCh. 7 - Prob. 7B.3BSTCh. 7 - Prob. 7B.4ASTCh. 7 - Prob. 7B.4BSTCh. 7 - Prob. 7B.5ASTCh. 7 - Prob. 7B.5BSTCh. 7 - Prob. 7B.1ECh. 7 - Prob. 7B.2ECh. 7 - Prob. 7B.3ECh. 7 - Prob. 7B.4ECh. 7 - Prob. 7B.5ECh. 7 - Prob. 7B.6ECh. 7 - Prob. 7B.7ECh. 7 - Prob. 7B.8ECh. 7 - Prob. 7B.9ECh. 7 - Prob. 7B.10ECh. 7 - Prob. 7B.13ECh. 7 - Prob. 7B.14ECh. 7 - Prob. 7B.15ECh. 7 - Prob. 7B.16ECh. 7 - Prob. 7B.17ECh. 7 - Prob. 7B.18ECh. 7 - Prob. 7B.19ECh. 7 - Prob. 7B.20ECh. 7 - Prob. 7B.21ECh. 7 - Prob. 7B.22ECh. 7 - Prob. 7C.1ASTCh. 7 - Prob. 7C.1BSTCh. 7 - Prob. 7C.2ASTCh. 7 - Prob. 7C.2BSTCh. 7 - Prob. 7C.1ECh. 7 - Prob. 7C.2ECh. 7 - Prob. 7C.3ECh. 7 - Prob. 7C.4ECh. 7 - Prob. 7C.5ECh. 7 - Prob. 7C.6ECh. 7 - Prob. 7C.7ECh. 7 - Prob. 7C.8ECh. 7 - Prob. 7C.9ECh. 7 - Prob. 7C.11ECh. 7 - Prob. 7C.12ECh. 7 - Prob. 7D.1ASTCh. 7 - Prob. 7D.1BSTCh. 7 - Prob. 7D.2ASTCh. 7 - Prob. 7D.2BSTCh. 7 - Prob. 7D.1ECh. 7 - Prob. 7D.2ECh. 7 - Prob. 7D.3ECh. 7 - Prob. 7D.5ECh. 7 - Prob. 7D.6ECh. 7 - Prob. 7D.7ECh. 7 - Prob. 7D.8ECh. 7 - Prob. 7E.1ASTCh. 7 - Prob. 7E.1BSTCh. 7 - Prob. 7E.1ECh. 7 - Prob. 7E.2ECh. 7 - Prob. 7E.3ECh. 7 - Prob. 7E.4ECh. 7 - Prob. 7E.5ECh. 7 - Prob. 7E.6ECh. 7 - Prob. 7E.7ECh. 7 - Prob. 7E.8ECh. 7 - Prob. 7E.9ECh. 7 - Prob. 1OCECh. 7 - Prob. 7.1ECh. 7 - Prob. 7.2ECh. 7 - Prob. 7.3ECh. 7 - Prob. 7.4ECh. 7 - Prob. 7.5ECh. 7 - Prob. 7.6ECh. 7 - Prob. 7.7ECh. 7 - Prob. 7.9ECh. 7 - Prob. 7.11ECh. 7 - Prob. 7.14ECh. 7 - Prob. 7.15ECh. 7 - Prob. 7.17ECh. 7 - Prob. 7.19ECh. 7 - Prob. 7.20ECh. 7 - Prob. 7.23ECh. 7 - Prob. 7.25ECh. 7 - Prob. 7.26ECh. 7 - Prob. 7.29ECh. 7 - Prob. 7.30ECh. 7 - Prob. 7.31E
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Kinetics: Initial Rates and Integrated Rate Laws; Author: Professor Dave Explains;https://www.youtube.com/watch?v=wYqQCojggyM;License: Standard YouTube License, CC-BY