Chemical thermodynamics Class 11 Questions And answers
Very Short Answer Questions
Question 1
What is the basis of Hess's law of heat summation?
Answer
Enthalpy is a state fuction and it is independent of the path followed.
Question 2
If enthalpy of fusion and enthalpy of vaporisation of sodium metal are 2.6 and 98.2 kJ/mol respectively, what is the enthalpy of sublimation of sodium?
Answer
Enthalpy of Sublimation = Enthalpy of fusion + Enthalpy of vaporisation = 2.6 +98.2 = 100.8 kJ/mol
Question 3
Can enthalpy of combustion be positive?
Answer
No, enthalpy of combustion is always negative.
Question 4
5 litre of an ideal gas at a pressure of 10atm expands isothermally into a vaccum until its volumne is 10 litre. How much heat is absorbed and work done in the expansion
Answer
We have $q=-w=P_{ext} \Delta V = 0$ As $P_{ext}=0$
Question 5
Which of the following is a state functions?
Work , heat,Volume, Enthalpy, Temperature, Pressure,Entropy
Answer
Volume, Enthalpy, Temperature, Pressure,Entropy
Question 6
Which of the following is extensive /intensive property
Volume, Enthalpy, Temperature, Density,Boiling point, Pressure, Specific heat
Answer
Extensive Property: Volume, Enthalpy
Intensive Property:Temperature, Density,Boiling point, Pressure, Specific heat
Question 7
What is the unit of Specific heat?
Answer
The units of specific heat are J/K gram (SI units)
Question 8
The standard enthalpies of formation of three substances A, B and C are 201.6, 52.6 and -106.4 k.J respectively. Arrange them in order of decreasing stability with respect to decomposition.
Answer
$C > B > A$
Question 9
What are the units of entropy and free energy?
Answer
J/K mol, J/mol
Question 10
Predict the entropy change sign in the following process
(i) Dissolution
(ii) Fusion
(iii)crystallization
Answer
(i) +
(ii) +
(iii) -
Short Answer Questions
Question 1
How can a reaction with positive changes in enthalpy and entropy be made entropy driven?
Answer
Endothermic spontanous reactions are entropy driven reaction
We know that
$\Delta G = \Delta H - T\Delta S$
Given $ \Delta H= \Delta S=+$
Now a reaction to be spontanous $\Delta G < 0$. This can be done in this case if $T \Delta S > \Delta H$
So We can make the reaction spontanous by increasing the temperature
Question 2
Heat capacity ($C_p$ ) is an extensive property but specific heat ($c$) is an intensive property. What will be the relation between $C_p$ and $c$ for 1 mol of water?
Answer
For water, heat capacity = 18 * specific heat
or $C_p = 18 \times c$
Specific heat = c = 4.18 J/K gram
Heat capacity = $C_p = 18 \times 4.18 = 75.3 J/k$
Question 3
Predict the sign of change in entropy (positive or negative) for the following changes:
(i) $HCOOH(l)-> H_2O(l) + CO_2(g)$
(ii)$NH_3(g) + HCl(g)-> NH_4Cl(s)$
(iii) $I_2(g)-> I_2(s)$
(iv) $20_3(g) -> 30_2(g)$
(v) $2H_2(g) + O_2(g) → 2H_2O(l)$
Answer
(i) +
(ii) -
(iii) +
(iv) +
(v) -
Question 4
Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
Answer
At constant volume
By first law of thermodynamics:
$q = \Delta U + (–w)$
$(–w) = p \Delta V$
Therefore
$q = \Delta U + p \Delta V$
Since volume is constant.
we have
$q_v = \Delta U$
So, $ q_v = \Delta U$ = change in internal energy
At constant pressure
$q_p = \Delta U + p \Delta V$
But, $\Delta U + p \Delta V = \Delta H$
Therefore
$q_p = \Delta H$ = change in enthalpy
Question 5
What is the state of a chemical reaction when
(i) $\Delta G=0$
(ii) $\Delta G < 0$
Answer
(i) Chemical reaction is in a state of equilibrium
(ii) Chemical reaction is spontanous
Question 6
State the first law of thermodynamics for (i) an adiabatic process (ii) a system undergoing a change in which internal energy remains constant
Answer
We know from first law of thermodynamics
$\Delta U = q + w$
(i) For Adiabatic process q=0, therefore $\Delta U =w$
The change in internal energy is equal to work done on the system
(ii) for $\Delta U =0$, q=-w
So heat absorbed by system is used in doing work
Question 7
Predict the sign of $\Delta G$ for a reaction that is
(a) Exothermic and accompanied by an increase in entropy.
(b) Endothermic and accompanied by increase in entropy.
(c) Can a temperature change affect the sign of $\Delta G$ in (a) or (b)? If so, how?
Answer
(a) -ve
(b) Cannot be predicted because it depends upon the magnitude of $\Delta S$, $\Delta H$ and T.
(c) A temperature change affects the sign in (b) by changing the magnitude of the temperature $-T \Delta S$
Question 8
Two litres of an ideal gas at 10 atm expands isothermally reversibly until its volume becomes 10 litre. How much heat is absorbed and work done in the expansion?
Question 1
Explain the following terms with examples
(i) Free energy
(ii) First law of thermodynamics
(iii)Enthalpy
(iv) Entropy
Question 2
Define the below terms
(i) Standard Enthalpy of Combustion
(ii) Standard Enthalpy of Formation of Substance
(iii) Bond Enthalpy
(iv) Enthalpy of Phase Change
Question 3
Derive the relationship between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.
Question 4
Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (pi , Vi ) to (pf , Vf ). With the help of a pV plot compare the work done in the above case with that carried out against a constant external pressure pf .
Numericals
Question 1
Determine the enthalpy of reaction for the hydrogenation pf acetylene to form ethane.
$C_2H_2(g) + 2H_2(g) -> C_2H_6(g)$
From the following given data
$2C_2H_2(g) + 5O_2(g) -> 4CO_2(g) + 2H_2O(l)$ , $\Delta H = -2600 kJ/Mol$
$2C_2H_6(g) + 7O_2(g) -> 4CO_2(g) + 6H_2O(l)$ , $\Delta H = -3120 kJ/Mol$
$H_2(g) + \frac {1}{2} O_2(g) -> H_2O(l)$ , $\Delta H = -286 kJ/Mol$
Answer
We have
$2C_2H_2(g) + 5O_2(g) -> 4CO_2(g) + 2H_2O(l)$ , $\Delta H = -2600 kJ/Mol$ -(i)
$2C_2H_6(g) + 7O_2(g) -> 4CO_2(g) + 6H_2O(l)$ , $\Delta H = -3120 kJ/Mol$ -(ii)
Reversing (ii), we have
$4CO_2(g) + 6H_2O(l) -> 2C_2H_6(g) + 7O_2(g)$ , $\Delta H = 3120 kJ/Mol$ --(iv)
Adding (i) ,(iv) and 4 (iii), we have
$C_2H_2(g) + 2H_2(g) =-> C_2H_6(g)$
Hence
$\Delta H = -2600 + 3120 - 4 \times 286=-624$ KJ/mol
Question 2
If enthalpies of formation for $C_2H 4(g)$,$CO_2(g)$ and $H_2O(l)$ at 25°C and 1 atm pressure are 52, -394 , - 286kJ / Mol respectively, then what is the enthalpy of combustion $C_2H_4$ ?
Question 3
Given:
$C + 2S -> CS_{2}$ ; $\Delta H = 117kJ$
$C + O_{2} -> CO_2$ , $\Delta H = - 393kJ$
$S + O_{2} -> SO_2$, $\Delta H = - 297kJ$
Find The heat of combustion of $CS_2$ to form $CO_{2}$ and $SO_{2}$
Answer
We have
$C + 2S -> CS_{2}$ ; $\Delta H = 117kJ$ -(i)
$C + O_{2} -> CO_2$ , $\Delta H = - 393kJ$ -(ii)
$S + O_{2} -> SO_2$, $\Delta H = - 297kJ$ -(iii)
Reversing equation(i), we get
$CS_{2} -> C + 2S$; $\Delta H = -117kJ$ -(iv)
Multiplying equation (iii) by 2, we get
$2S + 2O_{2} -> 2SO_2$, $\Delta H = - 594kJ$ -(v)
Adding (ii), (iv) and (v), we get
$CS_2 + 3 O_2 -> CO_2 + 2SO_2$
Hence
Enthalpy change = -393 -117 - 594 =-1104 KJ/mol
Question 4
$2Zn+O2 -> 2ZnO$, $\Delta G^0=-616 J$
$2Zn + S_2-> 2ZnS$, $\Delta G^0 = - 293J$
$S_2+20_2 -> 2S0_2$, $\Delta G^0 = - 408J$
Find the $\Delta G^0$ for the following
$2ZnS + 3O_2 -> 2ZnO +2SO_2$
Answer
Given
$2Zn+O2 -> 2ZnO$, $\Delta G^0=-616 J$ -(i)
$2Zn + S_2-> 2ZnS$, $\Delta G^0 = - 293J$ -(ii)
$S_2+20_2 -> 2S0_2$, $\Delta G^0 = - 408J$ -(iii)
Reversing reaction (ii), we get
$2ZnS -> 2Zn + S_2$, $\Delta G^0 = 293J$ -(iv)
Adding (ii) ,(iii) and (iv), we get
$2ZnS+ 3O_2 -> 2ZnO +2SO_2$
Hence
$\Delta G= 293 - 616 - 408=-731$ J
Question 5
The enthalpy of vaporization of mercury is 58.5 KJ/mol and the normal boiling point is 630K. What is the entropy of vaporization of mercury?
Question 6
The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.
$H_2(g) + Br_2(g) -> 2HBr(g)$
Given that Bond energy of $H_2$, $Br_2$ and HBr is 435 kJ/mol, 192 kJ /mol and 368 kJ /mol respectively.
Answer
Enthalpy Change= Bond energy of H2 + Bond energy of Br2 – 2 × Bond energy of HBr
= 435 + 192 – (2 × 368) kJ/mol
= –109 kJ /mol
Question 7
What is work done during the expansion of a gas from a volume of 4 dm3 to 6dm3 against a constant external pressure of 3 atm is (1 L atm= 101.32 J) ?
Question 8
Show that the reaction:
$CO(g) + \frac {1}{2} O_{2}(g)-> CO_2(g)$
at 300 K is spontaneous when standard entropy change is -0.094 kJ/ mol K. The standard Gibbs energies of formation of $CO_2$, and CO are -394.4 and -137.2 kJ/mol respectively.
Answer
$\Delta_{r} G ^0 =\Delta_{f} G^0 (products)- \Delta_{f} G^0 (reactants)$
$= - 394.4 - [- 137.2 + 0] = - 257.2 kJ/mol$
Since $\Delta_{r} G ^0 $ is negative, the reaction is spontaneous.
Now
$\Delta_{r} G ^0 = \Delta H^0 - T \Delta S$
$ - 257.2 = \Delta H^0 - 300 \times (- 0.094)$
$ \Delta H^0 = - 257.2 - 28.2 = - 285.4$ kJ/mol
Since $\Delta H$ is negative, the reaction is exothermic.
