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Chapter: 11th Chemistry : UNIT 7 : Thermodynamics

Solved Example Problem: Thermochemical Equations

Chemistry : Thermodynamics : Thermochemical Equations

Problem 

 

The standard enthalpies of formation of C2H5OH(l), CO2(g) and H2O(l) are - 277, -393.5 and -285.5 kJ mol-1 respectively.

Calculate the standard enthalpy change for the reaction 

C2H5OH(l)+3O2(g) →2CO2(g)+ 3 H2O(l) 

The enthalpy of formation of O2(g) in the standard state is Zero, by definition

 

Solution:

 

For example, the standard enthalpy change for the combustion of ethanol can be calculated from the standard enthalpies of formation of C2H5OH(l), CO2(g) and H2O(l). The enthalpies of formation are –277, – 393.5 and –285.5 kJ mol–1 respectively.

 

C3H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)


=[ −787 − 856 . 5] − [ −277]

= - 1643 .5 + 277

∆H0r = −1366 . 5 KJ

 


Problem 

Calculate the value of ∆U and ∆H on heating 128.0 g of oxygen from 0o C to 1000 C. CV and CP on an average are 21 and 29 J mol-1 K-1. (The difference is 8Jmol-1 K-1 which is approximately equal to R)

Solution.

We know

∆U   = n Cv (T2-T1)

∆H   = n CP (T2- T1)

Here

n= 128/32 4 moles ;


T2 = 1000

C =373K;

T1 = 00

C = 273K

∆U   = n Cv (T2-T1)

∆U =    4 x 21 x (373 - 273)

∆U =    8400 J

∆U =     8.4 kJ

∆H   = n Cp (T2- T1)

∆H =    4 × 29 × (373- 273)

∆H =    11600 J

∆H =    11.6 kJ

 

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11th Chemistry : UNIT 7 : Thermodynamics : Solved Example Problem: Thermochemical Equations |

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11th Chemistry : UNIT 7 : Thermodynamics


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