Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 5, Problem 5.123AP

(a)

Interpretation Introduction

Interpretation: The given reactions are to be identified as exothermic and endothermic.

The value of ΔHrxnο and the amount of energy absorbed or released is to be calculated for the given values. The ΔHrxnο for the metabolism of 1 mole of CH3OH(l) to give 1 mole of formaldehyde is to be compared with 509.8kJ .

Concept introduction: The reactions, in which heat energy is evolved, are termed as an exothermic process and if heat energy is absorbed then they are termed as an endothermic.

The standard enthalpy change (ΔHrxnο) is the change in enthalpy when one mole of product is formed at standard condition.

The amount of energy released or absorbed (q) is calculated by the formula,

q=n×ΔHrxnο

To determine: The given reaction is to be identified as exothermic and endothermic.

(a)

Expert Solution
Check Mark

Answer to Problem 5.123AP

Solution

The given reaction is exothermic.

Explanation of Solution

Explanation

The given reaction is stated as,

O2(g)+CH3OH(l)2HCOOH(l)+2H2O(l)+1019.6kJ

The reactions, in which heat energy is evolved, are termed as an exothermic process and if heat energy is absorbed then they are termed as an endothermic.

In the above reaction heat energy is evolved. Therefore, it is termed as an exothermic process.

(b)

Interpretation Introduction

To determine: The value of ΔHrxnο .

(b)

Expert Solution
Check Mark

Answer to Problem 5.123AP

Solution

The value of ΔHrxnο is -509.8kJ/mol_ .

Explanation of Solution

Explanation

The given reaction is stated as,

O2(g)+CH3OH(l)2HCOOH(l)+2H2O(l)+1019.6kJ

In the above reaction two moles of formic acid are formed. The standard enthalpy change (ΔHrxnο) is the change in enthalpy when one mole of product is formed at standard condition. The given value of evolved heat energy should be in negative (ΔHrxn,HCOOH=1019.6kJ) as the process is exothermic.

Hence, the ΔHrxnο of reaction is half of mole to the amount of evolved heat energy for the formation of formic acid (ΔHrxn,HCOOH) and calculated by the formula,

ΔHrxnο=ΔHrxn,HCOOH2mol (1)

Where,

  • ΔHrxnο is the standard enthalpy change for the formation of one mole of formic acid.
  • ΔHrxn,HCOOH is the given amount of evolved heat energy for the formation of formic acid.

Substitute the value of ΔHrxn,HCOOH in equation (1).

ΔHrxnο=1019.6kJ2mol=509.8kJ/mol

Hence, the value of ΔHrxnο is -509.8kJ/mol_ .

(c)

Interpretation Introduction

To determine: The amount of energy absorbed or released.

(c)

Expert Solution
Check Mark

Answer to Problem 5.123AP

Solution

The amount of energy released is -953.326kJ_ .

Explanation of Solution

Explanation

The mass (m) of methanol is 60.0g .

The given reaction is stated as,

O2(g)+CH3OH(l)2HCOOH(l)+2H2O(l)+1019.6kJ

The above reaction is metabolism of methanol liquid (CH3OH(l)) . The number of moles (n) used for metabolism is calculated by the formula,

n=mM (2)

Where,

  • m is the mass of methanol liquid.
  • M is the molar mass of methanol.

The molar mass of methanol is =1×C+4×H+1×O=1×12.011g/mol+4×1.008g/mol+1×15.999g/mol=12.011g/mol+4.032g/mol+15.999g/mol=32.042g/mol

Substitute the values of M and m in equation (2).

n=60.0g32.042g/mol=1.87mol

The amount of energy released or absorbed (q) is calculated by the formula,

q=n×ΔHrxnο (3)

Where,

  • ΔHrxnο is the standard enthalpy change for the formation of one mole of formic acid.

Substitute the values of ΔHrxnο and n   in equation (3).

q=1.87mol×509.8kJ/mol=953.326kJ

The negative sign indicates that the energy is released during the process. Hence, the amount of energy released is -953.326kJ_ .

(d)

Interpretation Introduction

To determine: The comparison of ΔHrxnο for the metabolism of 1 mole of CH3OH(l) to give 1 mole of formaldehyde with 509.8kJ .

(d)

Expert Solution
Check Mark

Answer to Problem 5.123AP

Solution

The ΔHrxnο for the metabolism of 1 mole of CH3OH(l) to give 1 mole of formaldehyde is expected to be smaller than 509.8kJ .

Explanation of Solution

Explanation

The amount of energy released in two step for the formation of formic acid is 953.326kJ . The formation of formaldehyde is the first step and metabolized by one mole of methanol. Therefore, the amount of energy released would be half of the 953.326kJ .

=953.326kJ2=476.663kJ

The negative sign of 953.326kJ is eliminated as it only indicates that heat is evolved.

Hence, the ΔHrxnο for the metabolism of 1 mole of CH3OH(l) to give 1 mole of formaldehyde is expected to be smaller than 509.8kJ .

Conclusion

  1. a. The given reaction is exothermic.
  2. b. The value of ΔHrxnο is -509.8kJ/mol_ .
  3. c. The amount of energy released is -953.326kJ_ .
  4. d. The ΔHrxnο for the metabolism of 1 mole of CH3OH(l) to give 1 mole of formaldehyde is expected to be smaller than 509.8kJ

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Chapter 5 Solutions

Chemistry

Ch. 5.5 - Prob. 11PECh. 5.6 - Prob. 12PECh. 5.6 - Prob. 13PECh. 5.7 - Prob. 14PECh. 5.7 - Prob. 15PECh. 5.7 - Prob. 16PECh. 5.8 - Prob. 17PECh. 5 - Prob. 5.1VPCh. 5 - Prob. 5.2VPCh. 5 - Prob. 5.3VPCh. 5 - Prob. 5.4VPCh. 5 - Prob. 5.5VPCh. 5 - Prob. 5.6VPCh. 5 - Prob. 5.7VPCh. 5 - Prob. 5.8VPCh. 5 - Prob. 5.9QPCh. 5 - Prob. 5.10QPCh. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Prob. 5.21QPCh. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Prob. 5.34QPCh. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Prob. 5.39QPCh. 5 - Prob. 5.40QPCh. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Prob. 5.43QPCh. 5 - Prob. 5.44QPCh. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - Prob. 5.49QPCh. 5 - Prob. 5.50QPCh. 5 - Prob. 5.51QPCh. 5 - Prob. 5.52QPCh. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - Prob. 5.58QPCh. 5 - Prob. 5.59QPCh. 5 - Prob. 5.60QPCh. 5 - Prob. 5.61QPCh. 5 - Prob. 5.62QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Prob. 5.76QPCh. 5 - Prob. 5.77QPCh. 5 - Prob. 5.78QPCh. 5 - Prob. 5.79QPCh. 5 - Prob. 5.80QPCh. 5 - Prob. 5.81QPCh. 5 - Prob. 5.82QPCh. 5 - Prob. 5.83QPCh. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - Prob. 5.88QPCh. 5 - Prob. 5.89QPCh. 5 - Prob. 5.90QPCh. 5 - Prob. 5.91QPCh. 5 - Prob. 5.92QPCh. 5 - Prob. 5.93QPCh. 5 - Prob. 5.94QPCh. 5 - Prob. 5.95QPCh. 5 - Prob. 5.96QPCh. 5 - Prob. 5.97QPCh. 5 - Prob. 5.98QPCh. 5 - Prob. 5.99QPCh. 5 - Prob. 5.100QPCh. 5 - Prob. 5.101QPCh. 5 - Prob. 5.102QPCh. 5 - Prob. 5.103APCh. 5 - Prob. 5.104APCh. 5 - Prob. 5.105APCh. 5 - Prob. 5.106APCh. 5 - Prob. 5.107APCh. 5 - Prob. 5.108APCh. 5 - Prob. 5.109APCh. 5 - Prob. 5.110APCh. 5 - Prob. 5.111APCh. 5 - Prob. 5.112APCh. 5 - Prob. 5.113APCh. 5 - Prob. 5.114APCh. 5 - Prob. 5.115APCh. 5 - Prob. 5.116APCh. 5 - Prob. 5.117APCh. 5 - Prob. 5.118APCh. 5 - Prob. 5.119APCh. 5 - Prob. 5.120APCh. 5 - Prob. 5.121APCh. 5 - Prob. 5.122APCh. 5 - Prob. 5.123APCh. 5 - Prob. 5.124APCh. 5 - Prob. 5.125APCh. 5 - Prob. 5.126APCh. 5 - Prob. 5.127APCh. 5 - Prob. 5.128APCh. 5 - Prob. 5.129APCh. 5 - Prob. 5.130APCh. 5 - Prob. 5.131APCh. 5 - Prob. 5.132APCh. 5 - Prob. 5.133APCh. 5 - Prob. 5.134APCh. 5 - Prob. 5.135APCh. 5 - Prob. 5.136APCh. 5 - Prob. 5.137APCh. 5 - Prob. 5.138APCh. 5 - Prob. 5.139APCh. 5 - Prob. 5.140APCh. 5 - Prob. 5.141APCh. 5 - Prob. 5.142APCh. 5 - Prob. 5.143AP
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