12th Grade > Chemistry
CHEMICAL THERMODYNAMICS MCQs
Total Questions : 30
| Page 3 of 3 pages
Answer: Option C. -> isolated system
:
C
There is no exchange of matter or energy from the flask.
:
C
There is no exchange of matter or energy from the flask.
Answer: Option C. -> 0 kJ
:
C
△H=nCp△Tsolution;since△T=0
so, △H=0
:
C
△H=nCp△Tsolution;since△T=0
so, △H=0
Answer: Option C. -> −900 J
:
C
w=−PΔV=−105(1×10−2−1×10−3)
= −900J
:
C
w=−PΔV=−105(1×10−2−1×10−3)
= −900J
Answer: Option B. -> 70 kcal
:
B
CH4 has 4 C - H bonds
∴ Energy needed to break a single C - H bond is 4004= 100 Kcal
C2H6 has 6 C - H bonds
∴ The C - C bond energy can be given by
670 Kcal - 6 × 100 Kcal = 70 Kcal
:
B
CH4 has 4 C - H bonds
∴ Energy needed to break a single C - H bond is 4004= 100 Kcal
C2H6 has 6 C - H bonds
∴ The C - C bond energy can be given by
670 Kcal - 6 × 100 Kcal = 70 Kcal
Answer: Option C. -> −16.11 kJ
:
C
The standard enthalpy of the combustion of glucose can be calculated by the equ.
C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
ΔHc=6×ΔHf(CO2)+6×ΔHf(H2O)−ΔHf[C6H12O6]
ΔH0=−2900180=−16.11kJ/gm
:
C
The standard enthalpy of the combustion of glucose can be calculated by the equ.
C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
ΔHc=6×ΔHf(CO2)+6×ΔHf(H2O)−ΔHf[C6H12O6]
ΔH0=−2900180=−16.11kJ/gm
Answer: Option C. -> -23.92 Cal
:
C
Given W=−140J;q=405
ΔE=q+W=40−140=−100J(1cal=4.18J)=−23.92Cal
:
C
Given W=−140J;q=405
ΔE=q+W=40−140=−100J(1cal=4.18J)=−23.92Cal
Answer: Option C. -> q=0
:
C
For adiabatic process that remain const i.e.q−0,△U=W
:
C
For adiabatic process that remain const i.e.q−0,△U=W
Answer: Option C. -> -101.19 K col
:
C
H(g) + O(g) → O - H(g) Required equation
Given: 12H2(g)+12O2(g)→OH(g);△H=+10.06Kcal
H(g)→12H2(g);△H=−52.09Kcal
O(g)→12O2(g);△H=−59.16Kcal
Adding them we get H(g)+O(g)+(OH)(g)△H=−101.19Kcal
:
C
H(g) + O(g) → O - H(g) Required equation
Given: 12H2(g)+12O2(g)→OH(g);△H=+10.06Kcal
H(g)→12H2(g);△H=−52.09Kcal
O(g)→12O2(g);△H=−59.16Kcal
Adding them we get H(g)+O(g)+(OH)(g)△H=−101.19Kcal
Answer: Option D. -> - 1.13 KJ
:
D
The given equations for the combustions of different types of phosphorous are:
(y) P4+5O2⟶P4O10−9.91KJ−−−−−−(i)
(R) P4+5O2⟶P4O10−8.78KJ−−−−−−(ii)
Now when we subtract we get (i) - (ii) gives
(y) P4⟶(R)P4−1.13KJ
:
D
The given equations for the combustions of different types of phosphorous are:
(y) P4+5O2⟶P4O10−9.91KJ−−−−−−(i)
(R) P4+5O2⟶P4O10−8.78KJ−−−−−−(ii)
Now when we subtract we get (i) - (ii) gives
(y) P4⟶(R)P4−1.13KJ
Answer: Option A. -> 57 × 34
:
A
H−+OH−→H2O+57kJmol−1
nH+=2×nH2SO4=2×0.5=1.0
nOH=nNaOH=0.75
Heat evolved = 0.75×57=34×57kJ
:
A
H−+OH−→H2O+57kJmol−1
nH+=2×nH2SO4=2×0.5=1.0
nOH=nNaOH=0.75
Heat evolved = 0.75×57=34×57kJ