Chemistry Chapter 3
Chemistry Chapter 3
Chemistry Chapter 3
Thermodynamics
MCQs.
1.
Which of the following is the symbolic form of the first law of
thermodynamics?
2.
(A) U = q W
(B) W = U q
(C) U = q + PV
(D) q + W = U
order
4.
5.
1
2
O2(g) MgO(S)
8.
9.
(B)
qrev
T
(c) qrev + T
(C) G = -nFEcell
12.
11.
(d) qrev T
n
F
(b) G = -
(d) G =
nF
E cell
13.
Ecell
(D) H = 0
15.
(B) 0
(D) Positive
(C) Negative
(D)
H = -ve S = -ve
18.
19.
e
RT
( )
-G / RT
(B) Kp =
G
RT
Answer:
1.
2.
3.
4.
5.
6.
7.
B
D
D
B
D
B
D
w=V-q
Extensive property
temp. can be of any order
Constant pressure
N2+O2 2NO(g)
Change of water into ice
enthalpy will be constant
qrev
8. B
T
9. B
10.
11.
12.
13.
H = -ve, S = +ve
C
G = -nFEcell
C
negative & positive
B
q=0
A
G may be less (av) more (ov) equal to H
Q1.
Q2.
Q3.
For example, the c6ld or hot water poured in the bottle and closed
with a cork.
Q4.
Q5.
Q6.
Q7.
Q8.
Q9.
Absorbed heat
Differencetemperature
V2
V1
= 2.303 R log
V2
V1
Where, V1
Where, V2
(sublimation)
H sublimation
T
Q16. The ice cube kept at room temperature melts on its own. Which law
of thermodynamics is obeyed by the reaction?
It obeys the first law of thermodynamics.
Q17. What is meant by temperature of a substance?
The level of thermal energy of a substance is called its temperature.
Q18. The entropy of water vapour is more than that of water. Why?
As the randomness in molecules is more in gaseous state than in
liquid state, the entropy of water vapour is more than that of water.
Q19. What will be the values of G at 270 K and 275 K?
At 270 K = -3c temperature, the ice does not melt on its own,
hence G is positive
At 275 K = 2c temperature, the ice will melt on its own, hence G
is negative.
Q20. Give definition of standard free energy of formation.
The value of change in free energy in direct reaction when 1 mole
compound is produced from its elements in standard state will be
equal to the value of change in standard free energy of formation of
the compound.
Q21. Write the equation showing relation between free energy change
and cell potential.
G = -nFE
ceil
= nRT
In
G = 2.303 nRT In
P1
P2
P1
P2
2.
3.
5.
7.
11.
qp
T
& qp = CP x T = H
H = U + (RT)
H = U + RT
Putting values of H & V from eqn.
1&2
CpT = CvT + RT
Cp =
T (Cv + R)
T
Cp = Cv + R
Cp Cv = R
- Similarly ratio of Cp & Cv is also constant Cp/Cv = r (gamma)
Where r=constant v=1.4 for ideal gas.
10.
q
T
Heat capacity =
Absorbed heat
Differencetemp
Specific
heat
capacity
Absorbed heat
( Differencetemperature) x(Weight of subs tancein gram)
Unit is joule/kelvin gram
Molar Heat Capacity:
The quantity of heat required for increase of temperature of I mole
substance by 1 C is called molar heat capacity.
Molar heat Capacity =
Absorbed heat
(Differencetemperature) x( Molecular weight)
12.
A system did the work of 785 joule after loss of 525 joule heat find
the chemical change in its internal energy.
The system losses heat
q = -525 joule
w = -785 joule
A system absorbed 650 joules heat & did the work its internal
energy change is 440 joules. Then find the out how much work has
been done.
The system absorbed heat
q = +650 joule
U = 440 joule
Where,
Now, G = -w (elect)
G = -nFE cell
It the cell is in standard state & the change in free energy is
associated with reaction & standard electrochemical cell potential,
than the relation between then can be shown by
G = -nFE cell
Potential
15.
Different
of
standard
red.
qrev = RT ln
qrev
T
= RT ln
V2
V 1 = 2.303 R log
S = R ln
qrev
T
V2
V1
V2
V1
= S
V2
V1
P1
P2
S = R ln
P1
P2
S = 2.303 R log
17.
P1
P2
19.
20.
* Long questions
1.
Explain internal energy and enthalpy and deduce the relation
between them.
* Internal Energy (U):
In every substance definite energy is stored in its on the basis of its
mass & characteristic structure, which is called internal energy.
It is stored in the form of kinetic energy & potential energy. The
compound particles of a substance have their linear rotation and
vibration energy. Their electron & nucleus also possess different
types of motion. So substance possesses kinetic energy.
Substance also-possesses potential energy due to repulsion
between electron-electron & repulsive forces between nucleusnucleus as well as attractive forces between electron & nucleus.
H = -56 KJ/mole
Write the Hells law of constant heat summation & explain giving
example. Mention its uses also.
Acco. Too Hesss law, "The total change in enthalpy in a chemical
reaction is equal to the algebraic sum of the changes in enthalpy
occurring during different steps."
This law can be explained on the basis of the first law of
thermodynamics.
The change in enthalpy at constant pressure and constant temp in a
chemical reaction is equal to the change in enthalpy H of that
reaction.
The enthalpy of reactants and products are constant at constant
pressure and constant tem. Hence, if the reaction occurs in one step
H = -393.5 KJ/mole
H(i) = -110.5 KJ/mole
H(ii) = -282.96 KJ/mole
(a) If two substances having different temp are kept in contact with
each other, the heat will always spontaneously flow from higher
temp to lower temp.
(b) Water always flows from higher level to lower level.
* Entropy:
Entropy is the measurement of randomness of the substance. To
understand consider following ex.
Suppose four squares of the size 1x1 and are drawn on a paper &
one honeybee flies over it.
1
2
1
2
3
3
4
4
5
6
7
8
9
Maximum & numbers 1, 2, 3, 4 will be required to show position of a
honeybee. Now, this honeybee flies over nine squares on the second
paper.
Maximum nine number 1 to 9 will be required to show the position
of the honeybee. Thus, more co-ordinates are required to show the
position of a honeybee.
So it can be said that if the honeybee flies & sits on the second
paper then, its entropy wile increases. The measurement of entropy
of any system is the measure of randomness of the system.
The entropy of crystalline solid substance is least because the
arrangement of molecules is systematic in crystalline solid.
While entropy of liquid substance is more than solid because of
randomness of molecule is more. Same way entropy of gaseous
substance is maximum, because the randomness of molecules in
the gas is maximum.
6.
R = 8.314 J.
n (g)
= np(g) n r (g)
= 4 (4+5)
= -5
H
= U + n(g) RT
The enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are -110
, -393.81, 82 and 9.7 kjoule mole -1, respectively. Calculate fH for
the following reaction: N2O4(g) + 3CO(g) 7 N2O(g) + 3CO2(g)
fH
3.
3
2
1
-1
2 O2(g) H2O(l) fH = - 286 kjoule mole
1
2
O2(g) CH3OH(2)
Now, reverse eqn. (1) multiply eqn. (3) by 2 and add eq n. (1) + eqn.
(2) + eqn. (3)
CO2(g) 2H2O(l) CH3OH(l)
C(graphite) + O2(g) CO2(g)
2H2(g) + O2(g)
2H2O(l)
3
2 O2
fH = +726 KJ/mole
fH = -393 KJ/mole
fH = -572 KJ/mole
C(graphite) + 2H2(g) +
4.
1
2
O2(g) CH3OH
fH = -239 KJ/mole
G = H - TS
0 = 400 T x 0.2
-0.2 T = -400
T=
400
0.2
= 2000K
For the following reaction, 2P(g) + Q(g) 2R(g) U = -10.5 kjoule and
S = -44.2 joulekelvin-1 Find G for the reaction. Will the reaction
occur on its own? Why?
2P + Q 2R
n(g)
= np(r) n r (g)
= 2 (2+1)
= -1
= U + n(g) RT
= -10.5 + (c-1 x 8.314 x 10-3 x 298)
= -10.5 2.477
= -12.977 KJ.
FG = H - TS
= -12.977 (298) (-44.2 x 10-3)
= 12.977 + 13.17
= 0.193 KJ/m
Here fG = +ve
rn will be not happen spontaneously.
6.
From the given data mention which of the following reaction will
occur on their own at 298 K.
Reaction: X : H = -52 kjoule; S = 956 joulekelvin-1
Reaction: Y : H = -60 kjoule; S = -65 joulekelvirr-1
x, y
for x G = H - TS
= -52 (298) (958 x 10-3)
= -52 285.4
= -337.4 KJ/m
G = -ve
It is spontaneous rn.
For y G = H - TS
= -60 (298) (-65 x 10-3)
= -60 (-19.5)
= -40.7 KJ/m
G = -ve spontaneous.
7.
= H - TS
= 170 (320) (26x10-3)
= 170 8.32
= 161.68 KJ/m
= +ve
not spontaneous
8.
1
2
NO and NO2 are 104.2 and 51.3 kjoule mole -1 respectively of 298 K
temperature.
No2(g) + O2(g) No2(g)
G
= GP - GR
= 51.3 - (104.2)
= -52.9 KJ
= -2.303 x RT log K
log K
=
Log k
52.9
2.308 x 8.314 x 1 03 x 298
= 9.2723
= 1.87x109
K
9.
G
2.303 x RT
Fe
= -nFE cell
= -2 x 96500 x 0.78
= -150540 J ( volt x couble = J)
150540
4.184
G = -2.303 RT log k
-35979.9 = -2.303 x 1.987 x 298 x log k
35779.9
2.303 x 1.987 x 298
log k =
= 26.3859
K = Anti (26.3859)
K = 2.432 x 1026
10.
The value of change in free energy for the following cell reaction at
298 K temperature is -76322 calories. Calculate the potential of the
electrochemical cell. F = 96500 coulomb, 1 Calorie = 4.184 joule.
Reaction: A(s) + B 2+( aq)
A 2+( aq)
+ B(s)
A + B+2 A+2+ B
G
E
= nFE
=
4
nF
1 calorie = 4.184 J
-76322C = ?
- 319331.2 J
E =
319331.2
2 x 96500
E = 1.65 v
11.
= -2.303 x RT log Kp