This document contains formulas and concepts related to thermodynamics. It defines formulas for changes in internal energy, work, enthalpy, entropy, Gibbs free energy, and reaction enthalpy and entropy. It also provides examples of numerical problems calculating thermodynamic quantities like heat, enthalpy change, entropy change, and temperature or reaction spontaneity based on given values and thermodynamic formulas.

1. 3day inservvice course

2. FORMULA THERMODYNAMIC
U=q+w
W=Pex.(-V)= - Pex V= Pex(Vf-Vi)
W=-∫Pin dV
Wrev=-∫ Pex dV =-∫ (PmC dp) dV
Wrev =-∫ Pm dV

3. Wrev =-∫ nRT dV/V=-nRT In Vf/Vi
=-2.303nRT log Vf/Vi
qp=(U2+pV2)-(U1+pV1)
H=U+pV
H= U + pV
PV= ngRT

4.  H= U + ngRT
 q=c x m x ∆T=CT
 H= U + RT
 Cp-Cv=R
 rH=(sum of enthalpies of product)-( sum
of enthalpies of reactant)
=∑ai H products-∑bi H reactants
i i
 rHº==∑ai jHºproducts-∑bi jHº reactants

5. rHº==∑ai jHºproducts-∑bi jHº
reactants
rH= rH1+ rH2+ rH3
rHv=∑ bond enthalpies reactants-∑
bond enthalpies products
S= qrev/T
Stotal= Ssystem+ Ssuit>0

6. G=H-TS
G= H- TS
 G= H- TS<0
rGº=-2.303RTlogK
rGº= rHº- T rSº=-RTInK

7. Numericals on thermodynamics
In a process,701J of heat is absorbed by a
system . what is the change in internal energy
for the process?
Sol. Heat absorbed by the system
(q)=+701J
Work done by the system (w)=-304J
Change in internal energy (∆U)=q+w=701-
394= 307J

8. The reaction of cyanamide, NH2CN with
oxygen was affected in a bomb
calorimeter and ∆U was found to be -
742.7KJ /mol of cyanamide at 298K.
Calculate the enthalpy change for the
reaction at 288K.
NH2CN(s)+3/2O2(g) N2(g)+CO2(g)+H2O
Sol. ∆U=-742.7KJ/mol;∆n(g)=2-3/2=+0.5
R=8.314*10-3kJ/K/mol;T=298.5K
∆H=-742.7+0.5*8.314*10-3 *298.15
=-742.7+1.239kJ=-741.15kJ

9.  Calculate the number of kJ necessary
to rise the temperature of 60ginium
from 35 to 55 C.
molar heat capacity of Al is
24J/mol/K
Sol. Moles of Al(n)=60/27=2.22mol
Molar heat capacity (Cm)=24J/mol/K
∆T=55-35=20K
q=Cm*n*∆T
=24*2.22*20= 1065.6J

10. Calculate the enthalpy change on freezing of
1.0 mol of water at -10 C to ice at-10C
.∆fus=6.03kJ /mol at 0 C. Cp[H2o(l)]=75.3J
/mol/k :Cp[H2o(s)]=36.8J /mol/K .
Sol. The freezing process is represented as
H2O(l) H2O(s) T1=-10 C =263.15K;
T2=0 C=273.15K;∆T=T2-T1=10K
Now according to Kirchoff’s equation ,
(∆H2-∆H1)/T2-T1=Cp (ice)-Cp (water)
(-6030-∆H1)/10=36.8-75.3
-6030-∆H1=10*38.5
∆H1= 5625J /mol

11. Enthalpy of combustion of carbon to carbon
dioxide is -3393.5kJ /mol . Calculate the
heat released upon formation of 35.2g of
Co2 from carbon to oxygen gas .
Sol. The combustion eq is
C(s) + O2 (g) Co2(g);∆cH=-393.5kJ/mol
Heat released in the formation of 44g of
Co2=393.5kJ
Heat released in the formation of 35.2g of
Co2=(393.5*35.2)/44=314.8kJ

12. The enthapy of formation of
Co(g),N2O(g) are -110,393,81 and 9.7kJ
/mol respectively. Find the value of ∆fH
for the reaction,N2O4(g)+3CO(g)
N2O(g)+3CO2(g).
Sol. ∆fH for the given reaction is given as
∆fH= ∑∆fH(products)-∑∆fH(reactants)
=∆fH (CO2)+ ∆fH (N2O) -3 ∆fH (CO)- ∆fH
(N2O4)
=3(-393)+81-3(-110)-9.7
=-1179+810+330-9.7= -777.7kJ /mol

13.  The eqilibrium constant fpr the vreaction is
10.caculate the value of ∆G;given
R=8J/K/mol;T=300K.
Sol. ∆G=-2.303RT log K
Here R =8J/K/Mol; T=300K ;K=10
∆G=-2.303*8*300*log 10
= -5527J/mol
Calculate the entropy change in surroundings
when 1 mol of H2o(l) formed under standard
conditions. Given ∆fH=-286kJ/mol .
Sol. ∆fH of H2O(l)=-286kJ/mol which refers to heat lost
to surrounds
q(surr)=- ∆fH of H2O(l)=-(-286)=+286kJ/mol
∆Ssurr=q(surr)/t=(+286*10³)/298.15= 959.24J/K/mol

14. Fir the reaction;2A(g)+B(g) 2D(g)
∆U298=-10.5kJ and ∆S=-44.1J
Calculate ∆G298 for the reaction and predict
whether the reaction is spontaneous or not.
Sol. ∆H= ∆U+ ∆n(g)RT
∆ n(g)=2- ∆ 3=-1mol;T=298K; U=10.5kJ
R=8.314kJ/K/mol
∆H=-10.5+[-1*8.314*10-³*298]
=12.298kJ
∆G= ∆H-T∆S
=-12.978-298*0.044
=+0.134kJ/mol
Since ∆G is positive, the reaction is spontaneous
∆

15. For a reaction at 298K; 2A+B C
H=400KJ/mol and ∆S=2kJ/mol.
At what temperaturewill be the reaction
become spontaneous considering ∆Hand ∆S
to be constant over the temperature range.
Sol. ∆G= ∆H-T∆S
At equilibrium, ∆G=0 ,
Therefore , ∆H=T∆ S
T=400/2=200K
Thus reaction will be in astate of equilibrium at
200K and will become spontaneous as the
temperature becomes>200K.