Class 11 ch 6 cbse medium unit-6-thermodynamics.ppt

this is about thermodynamics ppt

1. 3day inservvice course3day inservvice course

2. FORMULA THERMODYNAMICFORMULA THERMODYNAMIC
U=q+wU=q+w
W=PW=P
exex.(-.(-V)= - PV)= - P
ex ex V= PV= P
exex(V(V
ff-V-V
ii))
W=-∫PW=-∫P
inin dV dV
WW
revrev=-∫ P=-∫ P
exex dV =-∫ (P dV =-∫ (P
mmC dp) dVC dp) dV
WW
revrev =-∫ P =-∫ P
mm dV dV

3. WW
revrev =-∫ nRT dV/V=-nRT In V =-∫ nRT dV/V=-nRT In V
ff/V/V
ii
=-2.303nRT log V=-2.303nRT log V
ff/V/V
ii
qq
pp=(U=(U
22+pV+pV
22)-(U)-(U
11+pV+pV
11))
H=U+pVH=U+pV
H= H= U + pU + pVV
PPV= V= nn
ggRTRT

4. H= H= U + U + nn
ggRTRT
q=c x m x ∆T=Cq=c x m x ∆T=CTT
H= H= U + RU + RTT
CC
pp-C-C
vv=R=R
 
rrH=(sum of enthalpies of product)-( sum H=(sum of enthalpies of product)-( sum
of enthalpies of reactant)of enthalpies of reactant)
=∑a=∑a
i i H products-∑bH products-∑b
i i H reactantsH reactants
ii i i

rrHº==∑aHº==∑a
i i 
jjHºproducts-∑bHºproducts-∑b
i i 
jjHº reactantsHº reactants

5. 
rrHº==∑aHº==∑a
i i 
jjHºproducts-∑bHºproducts-∑b
i i 
jjHº Hº
reactantsreactants

rrH= H= 
rrHH
11+ + 
rrHH
22+ + 
rrHH
33

rrHH
vv
=∑ bond enthalpies =∑ bond enthalpies
reactantsreactants-∑ bond -∑ bond
enthalpies enthalpies
productsproducts
S= qS= q
revrev/T/T
SS
totaltotal= = SS
systemsystem+ + SS
suitsuit>0>0

6. G=H-TSG=H-TS
G= G= H- TH- TSS
 G= G= H- TH- TS<0S<0

rrGº=-2.303RTlogKGº=-2.303RTlogK

rrGº= Gº= 
rrHº- T Hº- T 
rrSº=-RTInKSº=-RTInK

7. Numericals o
n
Numericals o
n
thermody
namics
thermody
namics
In a process,701J of heat is absorbed by a In a process,701J of heat is absorbed by a
system . what is the change in internal energy system . what is the change in internal energy
for the process?for the process?
Sol. Sol. Heat absorbed by the system Heat absorbed by the system
(q)=+701J(q)=+701J
Work done by the system (w)=-304JWork done by the system (w)=-304J
Change in internal energy (Change in internal energy (∆U)=q+w=701-∆U)=q+w=701-
394= 394= 307J307J

8. The reaction of cyanamide, NHThe reaction of cyanamide, NH22CN with CN with
oxygen was affected in a bomb oxygen was affected in a bomb
calorimeter and calorimeter and ∆U was found to be -∆U was found to be -
742.7KJ /mol of cyanamide at 298K. 742.7KJ /mol of cyanamide at 298K.
Calculate the enthalpy change for the Calculate the enthalpy change for the
reaction at 288K.reaction at 288K.
NHNH22CN(s)+3/2O2(g) NCN(s)+3/2O2(g) N22(g)+CO(g)+CO22(g)+H(g)+H22OO
SolSol. ∆U=-742.7KJ/mol;∆n(g)=2-3/2=+0.5. ∆U=-742.7KJ/mol;∆n(g)=2-3/2=+0.5
R=8.314*10-3kJ/K/mol;T=298.5KR=8.314*10-3kJ/K/mol;T=298.5K
∆∆H=-742.7+0.5*8.314*10-3 *298.15H=-742.7+0.5*8.314*10-3 *298.15
=-742.7+1.239kJ=-742.7+1.239kJ=-741.15kJ=-741.15kJ

9. 
Calculate the number of kJ necessary Calculate the number of kJ necessary
to rise the temperature of 60ginium to rise the temperature of 60ginium
from 35 to 55 C. from 35 to 55 C.
molar heat capacity of Al is molar heat capacity of Al is
24J/mol/K24J/mol/K
SolSol. Moles of Al(n)=60/27=2.22mol . Moles of Al(n)=60/27=2.22mol
Molar heat capacity (Cm)=24J/mol/KMolar heat capacity (Cm)=24J/mol/K
∆∆T=55-35=20KT=55-35=20K
q=Cm*n*q=Cm*n*∆T∆T
=24*2.22*20= =24*2.22*20= 1065.6J1065.6J

10. Calculate the enthalpy change on freezing of Calculate the enthalpy change on freezing of
1.0 mol of water at -10 C to ice at-1.0 mol of water at -10 C to ice at-
10C .∆10C .∆fusfus=6.03kJ /mol at 0 C. =6.03kJ /mol at 0 C.
Cp[HCp[H22o(l)]=75.3J /mol/k :Cp[Ho(l)]=75.3J /mol/k :Cp[H22o(s)]=36.8J o(s)]=36.8J
/mol/K ./mol/K .
Sol.Sol. The freezing process is represented as The freezing process is represented as
HH22O(l) HO(l) H22O(s) T1=-10 C =263.15K;O(s) T1=-10 C =263.15K;
T2=0 C=273.15K;∆T=T2-T1=10K T2=0 C=273.15K;∆T=T2-T1=10K
Now according to Kirchoff’s equation ,Now according to Kirchoff’s equation ,
(∆H2-∆H1)/T2-T1=Cp (ice)-Cp (water)(∆H2-∆H1)/T2-T1=Cp (ice)-Cp (water)
(-6030-∆H1)/10=36.8-75.3(-6030-∆H1)/10=36.8-75.3
-6030-∆H1=10*38.5-6030-∆H1=10*38.5
∆ ∆H1= H1= 5625J /mol5625J /mol

11. Enthalpy of combustion of carbon to carbon Enthalpy of combustion of carbon to carbon
dioxide is -3393.5kJ /mol . Calculate the dioxide is -3393.5kJ /mol . Calculate the
heat released upon formation of 35.2g of heat released upon formation of 35.2g of
CoCo22 from carbon to oxygen gas from carbon to oxygen gas . .
SolSol. The combustion eq is. The combustion eq is
C(s) + OC(s) + O22 (g) Co (g) Co22(g);∆cH=-393.5kJ/mol(g);∆cH=-393.5kJ/mol
Heat released in the formation of 44g of Heat released in the formation of 44g of
CoCo22=393.5kJ=393.5kJ
Heat released in the formation of 35.2g of Heat released in the formation of 35.2g of
CoCo22=(393.5*35.2)/44==(393.5*35.2)/44=314.8kJ 314.8kJ

12. The enthapy of formation of The enthapy of formation of
Co(g),N2O(g) are -110,393,81 and Co(g),N2O(g) are -110,393,81 and
9.7kJ /mol respectively. Find the value of 9.7kJ /mol respectively. Find the value of
∆fH for the reaction,N∆fH for the reaction,N22OO44(g)+3CO(g) (g)+3CO(g)
N N22O(g)+3COO(g)+3CO22(g).(g).
Sol.Sol. ∆ ∆ffH for the given reaction is given asH for the given reaction is given as
∆∆ffH= H= ∑∑∆∆ffH(products)-H(products)-∑∑∆∆ffH(reactants)H(reactants)
=∆=∆ffH (CO2)+ ∆H (CO2)+ ∆ffH (N2O) -3 ∆H (N2O) -3 ∆ffH (CO)- H (CO)- ∆f∆fH H
(N2O4)(N2O4)
=3(-393)+81-3(-110)-9.7=3(-393)+81-3(-110)-9.7
=-1179+810+330-9.7= =-1179+810+330-9.7= -777.7kJ /mol-777.7kJ /mol

13. 
The eqilibrium constant fpr the vreaction is The eqilibrium constant fpr the vreaction is
10.caculate the value of 10.caculate the value of ∆G;given ∆G;given
R=8J/K/mol;T=300KR=8J/K/mol;T=300K..
Sol.Sol. ∆G=-2.303RT log K ∆G=-2.303RT log K
Here R =8J/K/Mol; T=300K ;K=10Here R =8J/K/Mol; T=300K ;K=10
∆∆G=-2.303*8*300*log 10G=-2.303*8*300*log 10
= = -5527J/mol-5527J/mol
Calculate the entropy change in surroundings Calculate the entropy change in surroundings
when 1 mol of H2o(l) formed under standard when 1 mol of H2o(l) formed under standard
conditions. Given ∆fH=-286kJ/mol .conditions. Given ∆fH=-286kJ/mol .
Sol.Sol. ∆ ∆ffH of HH of H22O(l)=-286kJ/mol which refers to heat lost O(l)=-286kJ/mol which refers to heat lost
to surrounds to surrounds
q(surr)=- ∆q(surr)=- ∆ffH of HH of H22O(l)=-(-286)=+286kJ/molO(l)=-(-286)=+286kJ/mol
∆∆Ssurr=q(surr)/t=(+286*10³)/298.15= Ssurr=q(surr)/t=(+286*10³)/298.15= 959.24J/K/mol959.24J/K/mol

14. Fir the reaction;2A(g)+B(g) 2D(g)Fir the reaction;2A(g)+B(g) 2D(g)
∆∆U298=-10.5kJ and ∆S=-44.1JU298=-10.5kJ and ∆S=-44.1J
Calculate ∆G298 for the reaction and predict Calculate ∆G298 for the reaction and predict
whether the reaction is spontaneous or notwhether the reaction is spontaneous or not..
Sol.Sol. ∆H= ∆U+ ∆n(g)RT ∆H= ∆U+ ∆n(g)RT
∆ ∆ n(g)=2- ∆ 3=-1mol;T=298K; U=10.5kJn(g)=2- ∆ 3=-1mol;T=298K; U=10.5kJ
R=8.314kJ/K/molR=8.314kJ/K/mol
∆∆H=-10.5+[-1*8.314*10-³*298]H=-10.5+[-1*8.314*10-³*298]
=12.298kJ=12.298kJ
∆∆G= ∆H-T∆SG= ∆H-T∆S
=-12.978-298*0.044=-12.978-298*0.044
=+0.134kJ/mol=+0.134kJ/mol
Since ∆G is positive, the reaction is spontaneousSince ∆G is positive, the reaction is spontaneous
∆∆

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