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CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 14 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Thermodynamics
Quiz 14
When two mole of an ideal gas
(
C
p
.
m
=
5
2
R
)
heated from
300
K
to
600
K
at constant pressure. The change in entropy of gas (
Δ
S
) is:
Report Question
0%
−
3
2
R
ln
2
0%
3
2
R
ln
2
0%
5
R
ln
2
0%
5
2
R
ln
2
Explanation
Solution:- (C)
5
R
ln
2
Δ
S
=
n
C
P
ln
T
2
T
1
+
n
R
ln
P
1
P
2
∵
Pressure is constant,
Δ
S
=
n
C
P
ln
T
2
T
1
Given:-
n
=
2
moles
C
P
=
5
2
R
T
2
=
600
K
T
1
=
300
K
∴
Δ
S
=
2
×
5
2
R
×
ln
600
300
⇒
Δ
S
=
5
R
ln
2
Hence the change in entropy of the gas is
5
R
ln
2
.
For the reaction
A
⇌
B
+
C
. At equilibrium, the concentration of
A
is
1
×
10
−
3
M
,
B
is
0.15
M
and
C
is
0.05
M
. The
△
G
o
for the reaction
A
at
27
o
C
will be
Report Question
0%
−
5
k
J
/
m
o
l
0%
−
17.3
k
J
/
m
o
l
0%
−
11.2
k
J
/
m
o
l
0%
n
o
n
e
o
f
t
h
e
s
e
Which of the following is/are correct ?
Report Question
0%
Δ
H
=
Δ
U
+
Δ
(PV) when P and V both changes
0%
Δ
H
=
Δ
U
+
P
Δ
V
when pressure is constant
0%
Δ
H
=
Δ
U
+
V
Δ
P
when volume is constant
0%
all are correct
Which of the following processes are spontaneous
Report Question
0%
Melting of ice at
2
atm and
273
K
0%
Melting point of ice at
1
/
2
atm and
273
K
0%
Boiling of water at
1
/
2
atm and
373
K
0%
Boiling of water at
2
atm and
373
K
For a reaction.
A
(
g
)
→
A
(
l
)
;
Δ
H
=
−
3
R
T
. The correct statement for the reaction is:
Report Question
0%
|
Δ
H
|
>
|
Δ
U
|
0%
Δ
H
=
Δ
U
≠
0
0%
|
Δ
H
|
<
|
Δ
U
|
0%
Δ
H
=
Δ
U
=
0
Explanation
Solution:- (A)
|
Δ
H
|
>
|
Δ
U
|
A
(
g
)
⟶
A
(
l
)
For the above reaction,
Δ
n
g
=
0
−
1
=
−
1
Now as we know that,
Δ
H
=
Δ
U
+
Δ
n
g
R
T
⇒
Δ
U
=
Δ
H
−
Δ
n
g
R
T
⇒
Δ
U
=
−
3
R
T
−
(
−
1
)
R
T
=
−
2
R
T
|
Δ
H
|
=
|
−
3
R
T
|
=
3
R
T
|
Δ
U
|
=
|
−
2
R
T
|
=
2
R
T
Hence
|
Δ
H
|
>
|
Δ
U
|
For the spontaneous process
2
F
(
g
)
→
F
2
(
g
)
, the sign of
Δ
H
and
Δ
S
respectively are?
Report Question
0%
+ve, -ve
0%
+ve, +ve
0%
-ve, -ve
0%
-ve, +ve
The favourable conditions for a spontaneous reaction are:
Report Question
0%
T
Δ
S
>
Δ
H
,
Δ
H
=
+
v
e
,
Δ
S
=
+
v
e
0%
T
Δ
S
>
Δ
H
,
Δ
H
=
+
v
e
,
Δ
S
=
−
v
e
0%
T
Δ
S
=
Δ
H
,
Δ
H
=
+
v
e
,
Δ
S
=
−
v
e
0%
T
Δ
S
=
Δ
H
,
Δ
H
=
+
v
e
,
Δ
S
=
+
v
e
Explanation
The free energy charge
△
G
is given by,
△
G
=
△
H
−
T
△
S
The reaction is spontaneous if
△
G
<
0.
⟹
△
H
−
T
△
S
<
0
⟹
T
△
S
>
△
H
,
△
H
=
+
v
e
,
△
S
=
+
v
e
For a reaction
R
1
,
Δ
G
=
x
K
J
m
o
l
−
1
. For a reaction
R
2
,
Δ
G
=
y
K
J
m
o
l
−
1
. Reaction
R
1
is non spontaneous but along with
R
2
it is spontaneous. This means that:
Report Question
0%
x
is
−
v
e
,
y
is
+
v
e
but in magnitude
x
>
y
0%
x
is
−
v
e
,
y
is
+
v
e
but in magnitude
y
>
x
0%
Both
x
and
y
are
−
v
e
but not equal
0%
Both
x
and
y
are
+
v
e
but not equal
Which of the following conditions make the process non spontaneous at all temperatures.
Report Question
0%
Δ
H
=
+
v
e
;
Δ
S
=
−
v
e
0%
Δ
H
=
−
v
e
;
Δ
S
=
+
v
e
0%
Δ
H
=
+
v
e
;
Δ
S
=
+
v
e
0%
Δ
H
=
−
v
e
;
Δ
S
=
−
v
e
Consider the formation of
M
g
O
(
s
)
. Assume that
△
H
o
and
△
S
o
are independent of temperature:
M
g
(
s
)
+
1
2
O
2
(
g
)
→
M
g
O
(
s
)
△
H
o
=
−
602
k
J
/
m
o
l
,
△
S
o
=
−
108
k
J
/
m
o
l
Calculate
△
G
for formation of
M
g
O
(
s
)
at
0
o
C
and is the reacion spontaneous or non spontaneous at
0
o
C
?
Report Question
0%
29
,
000
k
J
/
m
o
l
,
spontaneous
0%
−
29
,
000
k
J
/
m
o
l
,
non spontaneous
0%
−
30
,
000
k
J
/
m
o
l
,
spontaneous
0%
30
,
000
k
J
/
m
o
l
,
non spontaneous
For a spontaneous reaction, the
Δ
G
, equilibrium constant,
K
and will be respectively:
Report Question
0%
−
v
e
,
>
1
0%
+
v
e
,
1
0%
−
v
e
,
<
1
0%
−
v
e
.
1
For a gaseous reaction,
A
(
g
)
+
3
B
(
g
)
→
3
C
(
g
)
+
3
D
(
g
)
Δ
E
is
17
k
c
a
l
at
27
o
C
, assuming
R
=
2
c
a
l
K
−
1
m
o
l
−
1
, the value of
Δ
H
for the above reaction is:
Report Question
0%
15.8
k
c
a
l
0%
18.2
k
c
a
l
0%
20.0
k
c
a
l
0%
16.4
k
c
a
l
Δ
G
o
of
C
u
+
(
a
q
)
and
C
u
2
+
(
a
q
)
respectively are
+
50
and
+
66
k
J
/
m
o
l
e
. Value of (
Δ
H
o
−
T
Δ
S
o
) for
C
u
+
(
a
q
)
→
C
u
2
+
(
a
q
)
in kilo joules is?
Report Question
0%
−
16
0%
+
116
0%
−
116
0%
+
16
Which plot represents for an exothermic reaction:
Report Question
0%
0%
0%
0%
Explanation
For exothermic reaction:-
Δ
H
=
n
e
g
a
t
i
v
e
Δ
H
=
H
P
−
H
R
=
n
e
g
a
t
i
v
e
So,
H
R
>
H
P
(for exothermic reaction)
(Refer to Image)
This plot represents exothermic reaction.
Consider the reaction:
4
N
O
2
(
g
)
+
O
2
(
g
)
→
2
N
2
O
5
(
g
)
;
Δ
r
H
=
−
111
k
J
If
N
2
O
5
(
s
)
is formed instead of
N
2
O
5
(
g
)
in the above reaction, then
Δ
r
H
value will be:
[Given,
Δ
H
of sublimation for
N
2
O
5
is
54
k
J
m
o
l
−
1
]
Report Question
0%
+
54
k
J
0%
+
219
k
J
0%
−
219
k
J
0%
−
165
k
J
Explanation
4
N
O
2
(
g
)
+
O
2
(
g
)
⟶
2
N
2
O
5
(
g
)
;
Δ
H
=
−
111
k
J
.
.
.
.
.
(
1
)
N
2
O
5
(
s
)
⟶
N
2
O
5
(
g
)
;
Δ
H
=
54
k
J
N
2
O
5
(
g
)
⟶
N
2
O
5
(
s
)
;
Δ
H
=
−
54
k
J
2
×
[
N
2
O
5
(
g
)
⟶
N
2
O
5
(
s
)
;
Δ
H
=
−
54
k
J
]
2
N
2
O
5
(
g
)
⟶
2
N
2
O
5
(
s
)
;
Δ
H
=
−
108
k
J
.
.
.
.
.
(
2
)
Adding
e
q
n
(
1
)
&
(
2
)
, we have
4
N
O
2
(
g
)
+
O
2
(
g
)
+
2
N
2
O
5
(
g
)
⟶
2
N
2
O
5
(
g
)
+
2
N
2
O
5
(
s
)
;
Δ
H
=
[
(
−
111
)
+
(
−
108
)
]
k
J
4
N
O
2
(
g
)
+
O
2
(
g
)
⟶
2
N
2
O
5
(
s
)
;
Δ
H
=
−
219
k
J
Hence if
N
2
O
5
(
s
)
is formed instead of
N
2
O
5
(
g
)
, the value of
Δ
H
will be
−
219
k
J
.
Δ
S
s
u
r
r
o
u
n
d
i
n
g
s
=
+
959.1
J
K
−
1
m
o
l
−
1
Δ
S
s
y
s
t
e
m
=
−
163.1
J
K
−
1
m
o
l
−
1
. Then process is?
Report Question
0%
Spontaneous
0%
Non spontaneous
0%
At equilibrium
0%
Cannot be predicted from the information
Δ
S
for
4
F
e
(
s
)
+
3
O
2
(
s
)
→
2
F
e
2
O
3
(
s
)
is
−
550
J
/
m
o
l
/
K
. The process is found to be spontaneous even at
298
K
because (
Δ
H
=
−
1650
k
J
)
Report Question
0%
Δ
S
t
o
t
a
l
=
−
2000
J
0%
Δ
S
t
o
t
a
l
=
+
1650
J
0%
Δ
S
t
o
t
a
l
=
+
4980
J
0%
Δ
S
t
o
t
a
l
=
−
4980
J
In which of the following case entropy decreases-
Report Question
0%
Solid changing to liquid
0%
Expansion of a gas
0%
Crystals dissolve
0%
Polymerisation
Ethyl chloride (
C
2
H
5
C
l
) is prepared by reaction of ethylene with hydrogen cloride:
C
2
H
4
(
g
)
+
H
c
l
(
g
)
→
C
2
H
5
C
l
(
g
)
;
Δ
H
=
−
72.3
k
J
What is the value of
Δ
E
(in kJ), if
70
g
of ethylene and
73
g
of
H
C
l
are allowed to react at
300
K
Report Question
0%
−
69.8
0%
−
180.75
0%
−
174.5
0%
−
139.6
Consider the following processes
△
H
(
K
J
/
m
o
l
)
1
2
A
→
B
+
150
3
B
→
2
C
+
D
−
125
E
+
A
→
D
+
350
For
B
+
D
→
E
+
2
C
,
△
H
will be :
Report Question
0%
325
k
J
/
m
o
l
0%
525
k
J
/
m
o
l
0%
−
175
k
J
/
m
o
l
0%
−
325
k
J
/
m
o
l
An equilibrium reaction
X
+
Y
⇌
W
+
Z
,
△
H
=
+
v
e
is spontaneous in the forward direction. Then corresponding sign of
△
G
and
△
S
should be respectively:
Report Question
0%
+ve, -ve
0%
-ve , +ve
0%
+ve , +ve
0%
-ve , -ve
Rank the following substances in order of decreasing heat of combusion (maximum
→
minimum).
Report Question
0%
1
>
2
>
4
>
3
0%
3
>
4
>
2
>
1
0%
2
>
4
>
1
>
3
0%
1
>
3
>
2
>
4
In an insulated container 1 mole of a liquid. molar volume 100 ml at 1 bar. Liquid is steeply taken to 100 bar, when volume of liquid decreases by 1 ml. Find
Δ
H
for the process.
Report Question
0%
7900 bar mL
0%
8900 bar mL
0%
9900 bar mL
0%
10900 bar mL
the above reaction was carried out at 300 K in a bomb calorimeter. The heat released was 743 kJ/mol. The value of
△
H
300
K
for this reaction would be:
Report Question
0%
-740.5 kJ/mol
0%
-741.75 kJ/mol
0%
-743.0 kJ/mol
0%
-744.25 kJ/mol
For the reaction
A
⇌
B
+
C
At the equilibrium, the concentration of
A
is
1
×
10
−
3
M
is
B
is
0.15
M
and
C
is
0.05
M
. The
△
G
o
for the reaction at
27
o
C
will be
Report Question
0%
−
5
k
J
/
m
o
l
0%
−
17.36
k
J
/
m
o
l
0%
−
11.2
k
J
/
m
o
l
0%
10.1
k
J
/
m
o
l
Given the following data:
Substance
Δ
H
(KJ/mol)
S
∘
(J/mol K)
Δ
G
(KJ/mol)
FeO(s) -266.3 57.49 -245.12
C(Graphite) 0 5.74 0
Fe(s) 0 27.28 0
CO(g) -110.5 197.6 -137.15
Determine at what temperature the following reaction is spontaneous ?
FeO
(
s
)
+ C
(
Graphite
)
→
Fe
(
s
)
+ CO
(
g
)
Report Question
0%
289 K
0%
668 K
0%
966 K
0%
G
o
is
+ ve
. hence the reaction will never be spontaneous
For the process,
C
O
2
(
s
)
→
C
O
2
(
g
)
Report Question
0%
Both
Δ
H
and
Δ
S
are negative
0%
Δ
H
is negative and
Δ
S
is positive
0%
Δ
H
is +Ve and
Δ
S
is negative
0%
Both
Δ
H
and
Δ
S
are positive
Spontaneity may be observed in following conditions
Report Question
0%
△
H
= -ve,
△
S
= +ve
0%
△
H
= +ve,
△
S
= +ve
0%
△
H
= -ve,
△
S
= -ve
0%
All of these
Explanation
When the process is exothermic
(
△
H
s
y
s
t
e
m
<
0
)
,
and the entropy of the system increases
(
△
system
>
0
)
,
△
G
will be negative at
all temperatures. Thus, the process is always
spontaneous.
Using the data provided, calculate the multiple bond energy (kJ / mol) of a
C
=
C
bond in
C
2
H
2
. (Take the bond energy of a C-H bond as 350 kJ / mol)
2
C
(
s
)
+
H
2
(
g
)
→
C
2
H
2
(
g
)
Δ
H
=
225
k
J
/
m
o
l
2
C
(
s
)
→
2
C
(
g
)
;
Δ
H
=
1410
k
J
m
o
l
−
1
H
2
(
g
)
→
2
H
(
g
)
;
Δ
H
=
330
k
J
m
o
l
−
1
Report Question
0%
1165
0%
837
0%
865
0%
815
In a reaction, the change in entropy is given as 2.4 cal/K and the change in Gibbs free energy is given as 3.4 kcal, calculate the change in heat at the temperature of 20-degree centigrade?
Report Question
0%
3.4 kcal
0%
3.4 cal
0%
3.4 kJ
0%
3.4 J
For the reaction between
C
O
2
and graphite
C
O
2
(
g
)
+
C
(
s
)
→
2
C
O
(
g
)
Δ
H
=
170.0
K
J
a
n
d
Δ
s
=
170
J
K
−
1
. The reaction is spontaneous at
Report Question
0%
298 K
0%
500 K
0%
900 K
0%
none of the above
An endothermic reaction is spontaneous if
Report Question
0%
Δ
H
>
T
Δ
S
0%
Δ
H
<
T
Δ
S
0%
Δ
H
=
T
Δ
S
0%
T
Δ
S
=
0
For the reaction,
A
→
B
,
Δ
H
=
+
v
e
,
Δ
S
=
−
v
e
This reaction is
Report Question
0%
non-spontaneous at all temperature
0%
non-spontaneous at low temperature
0%
non-spontaneous at high temperature
0%
spontaneous at high temperature
For the reaction
X
2
Y
4
(
l
)
→
2
X
Y
2
(
g
)
at 300 K the values of
Δ
U
and
Δ
S
are
2
k
c
a
l
and
20
c
a
l
K
−
1
respectively. The value of
Δ
G
for the reaction is:
Report Question
0%
-3400 cal
0%
3400 cal
0%
2000 cal
0%
-2800 cal
The reaction
2
A
(
g
)
→
A
2
(
g
)
, will be spontaneous
Report Question
0%
At high temperature
0%
At low temperature
0%
At all temperature
0%
Never at any temperature
The correct thermodynamic conditions for the spontaneous reaction at all temperature is:
Report Question
0%
Δ
H
<
0
and
Δ
S
=
0
0%
Δ
H
>
0
and
Δ
S
<
0
0%
Δ
H
<
0
and
Δ
S
>
0
0%
Δ
H
<
0
and
Δ
S
<
0
Consider the following processes:
1
2
A
⟶
B
:
△
H
=
150
3
B
⟶
2
C
+
D
:
△
H
=
−
125
E
+
A
⟶
2
D
△
=
350
For B+D
→
E+2C,
△
H will be:
Report Question
0%
325 kJ/mol
0%
525 kJ/mol
0%
-175 kJ/mol
0%
-325 kJ/mol
The difference between heat of reaction at pressure and constant volume for the reaction,
C
H
2
=
C
H
2
(
g
)
+
3
O
2
(
g
)
→
2
C
O
2
(
g
)
+
2
H
2
O
(
l
)
at 300 K is?
Report Question
0%
5.87 kJ
0%
-4.99 kJ
0%
6.89 kJ
0%
-7.25 kJ
Given that
Z
n
+
1
/
2
O
2
→
Z
n
O
+
35.25
k
J
.
H
g
O
→
H
g
+
1
/
2
O
2
+
9.11
k
J
.
The heat of the reaction
Z
n
+
H
g
O
→
Z
n
O
+
H
g
is
Report Question
0%
-26.14kJ
0%
44.39kJ
0%
-44.39kJ
0%
26.14kJ
Explanation
Z
n
+
1
2
O
2
→
Z
n
O
+
35.25
k
J
H
g
O
→
H
g
+
1
2
O
2
+
9.11
k
J
By Adding both we get,
Z
n
+
H
g
O
→
Z
n
O
+
H
g
+
44.36
k
J
Heat of reaction is
−
44.39
k
J
What is the free energy change
△
G
, when
1.0
mole of water at
100
o
C
and
1
atm pressure is converted steam at
100
o
C
and
1
atm pressure:-
Report Question
0%
+540 cal
0%
-9800 cal
0%
+9800 cal
0%
0 cal
Explanation
As we know that
Δ
G
at equilibrium is always
0
. So, here also is equilibrium ; so
Δ
G
=
0
500 J of heat was supplied to a system at constant volume.It resulted in the increase of temperature of the system from
20
o
C
to
25
o
C
.What is the change in internal energy of the system?
Report Question
0%
+
8.43
J
0%
+
43
J
0%
+
458.43
J
0%
+
580.43
J
The P - V diagram of a system undergoing a thermodynamic process is shown in figure. Work done by the system in going from
A
→
B
→
C
is 30 J and 40 J heat is given to the system. The change in internal energy of the gas if the gas is directly taken from A to C is
Report Question
0%
10 J
0%
70 J
0%
84 J
0%
134 J
Explanation
For process
A
→
B
→
C
Δ
u
=
Δ
θ
−
Δ
ω
=
40
−
30
=
10
J
A
→
C
Δ
u
=
10
J
Hence
,
option
(
A
)
is correct
.
2
A
l
2
(
s
)
⟶
4
A
l
(
s
)
+
3
O
2
(
g
)
,
△
G
=
+
138
kcal considering the contributing of entropy to the spontaneity of this reaction, the reaction is________And entropy of the system_______.
Report Question
0%
Spontaneous increase
0%
Spontaneous decrease
0%
non-Spontaneous, increase
0%
non-Spontaneous decrease
For reaction
A
→
B
,
Δ
H
and
Δ
S
are positive. the most favourable condition of spontaneous process.
Report Question
0%
low temperature
0%
high temperature
0%
high concentration
0%
very low temperature
The Gibb's energy for the decomposition of
A
l
2
O
3
at
500
o
C
is as follows:
2
/
3
A
l
2
O
3
⟶
4
/
3
A
l
+
O
2
△
r
G
=
+
966
k
J
m
o
l
−
1
The potential difference needed for electrolytic reduction of
A
l
2
O
3
at
500
o
C
is at least
Report Question
0%
5.0 V
0%
4.5 V
0%
3.0 V
0%
2.5 V
Dextrorotatory
α
-pinene has a specific rotation
[
α
]
20
D
=
+
51.3
∘
.
A sample of
α
−
p
i
n
e
n
e
containing both the enantiomers was found to have a specific rotation value
[
α
]
20
D
=
+
30.8
∘
.
The percentages of the
(
+
)
and
(
−
)
enantiomers present in the sample are, respectively.
Report Question
0%
70
%
and
30
%
0%
80
%
and
20
%
0%
20
%
and
80
%
0%
60
%
and
40
%
N
H
3
(
g
)
+
3
C
l
2
→
←
\vbox
t
o
.5
e
x
\vss
N
C
l
4
(
g
)
+
3
H
C
l
:
−
Δ
H
1
N
2
(
g
)
+
3
H
2
(
g
)
→
←
\vbox
t
o
.5
e
x
\vss
2
N
H
3
(
g
)
:
Δ
H
2
H
2
(
g
)
+
C
l
2
(
g
)
→
←
\vbox
t
o
.5
e
x
\vss
2
H
C
l
(
g
)
:
Δ
H
3
The heat of formation of
N
C
l
4
in the terms of
Δ
H
1
,
Δ
H
2
,
Δ
H
3
is:
Report Question
0%
Δ
H
1
=
−
Δ
H
1
+
Δ
H
2
2
−
3
2
Δ
H
3
0%
Δ
H
1
=
Δ
H
1
+
Δ
H
2
2
−
3
2
Δ
H
3
0%
Δ
H
1
=
Δ
H
1
−
Δ
H
2
2
+
3
2
Δ
H
3
0%
none of these
An ideal gas expands according to the law
P
2
V
= constant. The internal energy of the gas
Report Question
0%
Increases continuously
0%
Decreases continuously
0%
Remain constant
0%
First increases and then decreases
In which of the following ionic compounds,
Δ
H
f
is negative only due to lattice energy:
(i)
N
a
F
(ii)
M
g
O
(iii)
L
i
2
N
(iv)
N
a
2
S
Report Question
0%
Only (iv)
0%
Only (iii) and (iv)
0%
Only (ii), (iii), (iv)
0%
All of these
When a bottle of perfume is opened, odorous molecules mix with air and slowly diffuse throughout the entire room. The incorrect fact about the process is:
Report Question
0%
△
G
=
−
V
e
0%
△
H
≃
0
0%
△
S
=
−
V
e
0%
△
S
=
+
v
e
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0
Answered
1
Not Answered
49
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Correct : 0
Incorrect : 0
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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