EBK CHEMISTRY
8th Edition
ISBN: 9780135216972
Author: Robinson
Publisher: PEARSON CO
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Chapter 15, Problem 15.139SP
Interpretation Introduction
Interpretation:
The rate constant for the reverse reaction at
Concept introduction:
For the equilibrium reaction, the expression for the equilibrium constant is the ratio of concentration of product to reactant raised to their
The equilibrium constant can also be calculated in terms of partial pressure of each gas at equilibrium. It is denoted by symbol Kpand equal to the ratio of partial pressure of product to reactant.
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Students have asked these similar questions
The rate constant of the elementary reaction
BrO(g) + NO(g) – Br(g) + NO2(g)
is 1.3 x 1010 L mol-1 s-1 at 25°C, and its equilibrium
constant is 5.0 x 1010 at this temperature. Calculate the
rate constant at 25°C of the elementary reaction
Br(g) + NO, (g) – BrO(g) + NO(g)
Consider the elementary reactions and their rate constants.
kf = 4.99 × 10–2s-1
k, = 3.67 × 10-' s-'
A(g) + B(g) → C(g) + D(g)
C(g) + D(g)
A(g) + B(g)
What is the equilibrium constant (K) of the following generic reaction?
A(g) + B(g) = C(g) + D(g)
Ke =
Which statement correctly describes the partial pressures of the reactants and products at equilibrium?
The partial pressures of the reactants are greater than the partial pressures of the products at equilibrium.
The partial pressures of the products are greater than the partial pressures of the reactants at equilibrium.
The partial pressures of the reactants and products are equal at equilibrium.
12)
At 25 °C, a certain elementary reaction has a forward rate constant
equal 1.0 x 10-³ s' and an equilibrium constant, equal to 4.19. What
is the rate constant for the reverse reaction?
Chapter 15 Solutions
EBK CHEMISTRY
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Prob. 15.57SPCh. 15 - Prob. 15.58SPCh. 15 - Prob. 15.59SPCh. 15 - For each of the following equilibria, write the...Ch. 15 - Prob. 15.61SPCh. 15 - The reaction 2AsH3(g)As2(g)+3H2(g) has Kp = 7.2107...Ch. 15 - Prob. 15.63SPCh. 15 - Calculate the value of the equilibrium constant at...Ch. 15 - Prob. 15.65SPCh. 15 - An equilibrium mixture of PCI5, PCi3, and Cl2 at a...Ch. 15 - The partial pressures in an equilibrium mixture of...Ch. 15 - At 298 K, Kc is 2.2105 for the reaction F (g) + O2...Ch. 15 - At 298 K, Kp is 1.6106 for the reaction...Ch. 15 - Prob. 15.70SPCh. 15 - Prob. 15.71SPCh. 15 - Prob. 15.72SPCh. 15 - Prob. 15.73SPCh. 15 - For each of the following equilibria, write the...Ch. 15 - For each of the following equilibria, write the...Ch. 15 - When the following reactions come to equilibrium,...Ch. 15 - Prob. 15.77SPCh. 15 - A chemical engineer is studying reactions to...Ch. 15 - Prob. 15.79SPCh. 15 - At 1400 K, Kc = 103 for the reaction...Ch. 15 - The first step in the industrial synthesis of...Ch. 15 - 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Prob. 15.134SPCh. 15 - Prob. 15.135SPCh. 15 - Prob. 15.136SPCh. 15 - Prob. 15.137SPCh. 15 - Prob. 15.138SPCh. 15 - Prob. 15.139SPCh. 15 - Prob. 15.140SPCh. 15 - Prob. 15.141SPCh. 15 - Prob. 15.142SPCh. 15 - Prob. 15.143SPCh. 15 - Prob. 15.144MPCh. 15 - Prob. 15.145MPCh. 15 - Refining petroleum involves cracking large...Ch. 15 - At 1000 K, Kp = 2.1 106 and H ° = -107.7 kJ for...Ch. 15 - Consider the gas-phase decomposition of...Ch. 15 - Prob. 15.149MPCh. 15 - Prob. 15.150MPCh. 15 - Prob. 15.151MPCh. 15 - Prob. 15.152MPCh. 15 - Prob. 15.153MPCh. 15 - Prob. 15.154MPCh. 15 - A 125.4 g quantity of water and an equal molar...Ch. 15 - Prob. 15.156MPCh. 15 - Prob. 15.157MPCh. 15 - Prob. 15.158MPCh. 15 - Prob. 15.159MPCh. 15 - Ozone is unstable with respect to decomposition to...Ch. 15 - Prob. 15.161MPCh. 15 - For the decomposition reaction...Ch. 15 - Prob. 15.163MP
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