General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 14, Problem 14.111SP

(a)

Interpretation Introduction

Interpretation:

The initial rate of formation of HI has to be determined.

Concept introduction:

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

  • The change in concentration term is divided by the respective stoichiometric coefficient.
  • The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
  • Rate of reaction is always represented by positive quantities.

(a)

Expert Solution
Check Mark

Answer to Problem 14.111SP

The initial rate of formation of HI is 1.13×103M/min

Explanation of Solution

From the given information, the units of rate constant denotes a second-order reaction and the rate law is probably as follows

rate = k[H2][I2]

By using the ideal gas equation to solve for the initial concentrations of H2andI2.  Then to calculate initial rate with respect to H2andI2 and then convert it into the initial rate of formation of HI and carryout the significant figure throughout this calculation to minimize rounding errors

n=PVRT

nV=M=PRT

The total pressure is 1658 mmHg and there are equimolar amounts of H2andI2 in the vessel and the partial pressure of each gas could be 829 mmHg.

[H2]=[I2]=(829mmHg×1atm760mmHg)(0.0821L.atmK.atm)(400+273)K=0.01974M

Now, let’s convert the units of the rate constant to /M.min, and substitute the corresponding values in the rate law to solve the rate as follows.

k = 2.42×10-21M.s×60s1min=1.4521M.min

rate = k[H2][I2]

rate =( 1.4521M.min)(0.01974M)(0.01974)=5.658×104M/min

We know that,

rate=12Δ[HI]Δt

or

Δ[HI]Δt=2×rate = (2)(5.658×104M/min)=1.13×103M/min

(b)

Interpretation Introduction

Interpretation:

The rate formation of HI and the concentration of HI in molarity after 10 min has to be determined.

Concept introduction:

Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants).  It is also said to be as rate equation.

Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.

  • The change in concentration term is divided by the respective stoichiometric coefficient.
  • The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
  • Rate of reaction is always represented by positive quantities.

(b)

Expert Solution
Check Mark

Answer to Problem 14.111SP

The rate formation of HI and the concentration of HI in molarity after 10 min is 8.80×103M

Explanation of Solution

The integrated second-order rate law is used to calculate the concentration of H2 after 10.0min.  Then substitute the obtained concentration into the rate law to get rate.

1[H2]t=(1.4521M.min)(10.0min)+10.01974M[H2]t=0.01534M

Similarly we can calculate the concentration of I2 after 10.0min and it will also equal to

[I2]t= 0.01534M

Now,

rate = k[H2][I2]

rate = ( 1.4521M.min)(0.01534M)(0.01534M)=3.417×104M/min

We know that,

Δ[HI]Δt= 2 × rate =(2)(3.417×104M/min)=6.83×104M/min

Therefore, the concentration of HI after 10.0 min is

[HI]t = ([H2]0- [H2]t×2

[HI]t =(0.01974 M- 0.01534 M)×2=8.80×103M

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Chapter 14 Solutions

General Chemistry

Ch. 14.4 - Practice Exercise The first-order rate constant...Ch. 14.4 - Review of Concepts (a) What can you deduce about...Ch. 14.5 - Practice Exercise The reaction between NO2 and CO...Ch. 14.5 - Prob. 1RCCh. 14.6 - Prob. 1RCCh. 14 - Prob. 14.1QPCh. 14 - 15.2 Explain the difference between physical...Ch. 14 - Prob. 14.3QPCh. 14 - Prob. 14.4QPCh. 14 - Prob. 14.5QPCh. 14 - 14.6 Consider the reaction Suppose that at a...Ch. 14 - Prob. 14.7QPCh. 14 - 14.8 What are the units for the rate constants of...Ch. 14 - Prob. 14.9QPCh. 14 - Prob. 14.10QPCh. 14 - Prob. 14.11QPCh. 14 - Prob. 14.13QPCh. 14 - Prob. 14.14QPCh. 14 - Prob. 14.15QPCh. 14 - Prob. 14.16QPCh. 14 - Prob. 14.17QPCh. 14 - Prob. 14.18QPCh. 14 - Prob. 14.19QPCh. 14 - Prob. 14.20QPCh. 14 - 14.21 What is the half-life of a compound if 75...Ch. 14 - 14.22 The thermal decomposition of phosphine (PH3)...Ch. 14 - Prob. 14.23QPCh. 14 - Prob. 14.24QPCh. 14 - 14.25 Consider the first-order reaction A → B...Ch. 14 - Prob. 14.26QPCh. 14 - 14.27 Define activation energy. What role does...Ch. 14 - Prob. 14.28QPCh. 14 - Prob. 14.29QPCh. 14 - 14.30 As we know, methane burns readily in oxygen...Ch. 14 - Prob. 14.31QPCh. 14 - Prob. 14.32QPCh. 14 - Prob. 14.33QPCh. 14 - Prob. 14.34QPCh. 14 - Prob. 14.35QPCh. 14 - Prob. 14.36QPCh. 14 - Prob. 14.37QPCh. 14 - 14.38 The rate at which tree crickets chirp is 2.0...Ch. 14 - 14.39 The diagram here describes the initial state...Ch. 14 - Prob. 14.40QPCh. 14 - Prob. 14.41QPCh. 14 - Prob. 14.42QPCh. 14 - 14.43 Explain why termolecular reactions are...Ch. 14 - 14.44 What is the rate-determining step of a...Ch. 14 - Prob. 14.45QPCh. 14 - Prob. 14.46QPCh. 14 - Prob. 14.47QPCh. 14 - Prob. 14.48QPCh. 14 - Prob. 14.49QPCh. 14 - Prob. 14.50QPCh. 14 - Prob. 14.51QPCh. 14 - Prob. 14.52QPCh. 14 - Prob. 14.53QPCh. 14 - Prob. 14.54QPCh. 14 - Prob. 14.55QPCh. 14 - Prob. 14.56QPCh. 14 - Prob. 14.57QPCh. 14 - Prob. 14.58QPCh. 14 - Prob. 14.59QPCh. 14 - Prob. 14.60QPCh. 14 - Prob. 14.61QPCh. 14 - Prob. 14.62QPCh. 14 - Prob. 14.63QPCh. 14 - Prob. 14.64QPCh. 14 - Prob. 14.65QPCh. 14 - 14.66 The decomposition of N2O to N2 and O2 is a...Ch. 14 - Prob. 14.67QPCh. 14 - Prob. 14.68QPCh. 14 - 14.69 Consider the zero-order reaction a → B....Ch. 14 - Prob. 14.70QPCh. 14 - Prob. 14.72QPCh. 14 - Prob. 14.73QPCh. 14 - Prob. 14.74QPCh. 14 - Prob. 14.75QPCh. 14 - Prob. 14.76QPCh. 14 - Prob. 14.77QPCh. 14 - Prob. 14.78QPCh. 14 - Prob. 14.79QPCh. 14 - Prob. 14.80QPCh. 14 - Prob. 14.81QPCh. 14 - Prob. 14.82QPCh. 14 - 14.83 When a mixture of methane and bromine is...Ch. 14 - 14.84 Consider this elementary step: (a)...Ch. 14 - Prob. 14.85QPCh. 14 - Prob. 14.86QPCh. 14 - 14.87 In recent years ozone in the stratosphere...Ch. 14 - Prob. 14.88QPCh. 14 - Prob. 14.90QPCh. 14 - Prob. 14.91QPCh. 14 - Prob. 14.92QPCh. 14 - Prob. 14.93QPCh. 14 - Prob. 14.94QPCh. 14 - Prob. 14.95QPCh. 14 - Prob. 14.96QPCh. 14 - Prob. 14.97QPCh. 14 - Prob. 14.98QPCh. 14 - Prob. 14.100QPCh. 14 - Prob. 14.101QPCh. 14 - 14.102 Consider the potential energy profiles...Ch. 14 - Prob. 14.103QPCh. 14 - Prob. 14.104QPCh. 14 - 14.105 The activation energy for the...Ch. 14 - Prob. 14.106QPCh. 14 - Prob. 14.107SPCh. 14 - Prob. 14.108SPCh. 14 - Prob. 14.109SPCh. 14 - Prob. 14.110SPCh. 14 - Prob. 14.111SPCh. 14 - Prob. 14.112SPCh. 14 - Prob. 14.113SPCh. 14 - Prob. 14.114SPCh. 14 - Prob. 14.115SPCh. 14 - 14.116 To prevent brain damage, a drastic medical...Ch. 14 - Prob. 14.117SPCh. 14 - Prob. 14.118SP
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