Class 11 Chemistry · 62 questions · 61 with answers
Thermodynamics is not concerned about______.
(iii) the rate at which a reaction proceeds.
Which of the following statements is correct?
(iii) The presence of reactants in a closed vessel made up of copper is an example of a closed system.
The state of a gas can be described by quoting the relationship between___.
(iv) pressure, volume, temperature, amount
The volume of gas is reduced to half from its original volume. The specific heat will be ______.
(iii) remain constant
During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is –1
(iii) C4H10(g) + O (g) → 4CO2 (g) + 5H2O ( l ) ∆cH = –2658.0 kJ mol 2 2 13 –1
∆f U of formation of CH4 (g) at certain temperature is –393 kJ mol–1. The value of ∆f H is
(ii) < ∆f U
In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.
(iii) q = 0, ∆T = 0, w = 0
(i) q = 0, ∆T ≠ 0, w = 0
The pressure-volume work for an ideal gas can be calculated by using the expression w = − ∫ pex dV . The work can also be calculated from the pV– plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume Vi to Vf . choose the correct option.
(ii) w (reversible) < w (irreversible)
(i) w (reversible) = w (irreversible)
The entropy change can be calculated by using the expression ∆S = . When water freezes in a glass beaker, choose the correct statement amongst the following :
(iii) ∆S (system) decreases but ∆S (surroundings) increases.
On the basis of thermochemical equations (a), (b) and (c), find out which of the algebric relationships given in options
(iii) x=y+z (iv) y = 2z – x
Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options
(iii) x>y (iv) x<y
(i) to
(ii) x = 2y
The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
(iii) may be positive or negative
Enthalpy of sublimation of a substance is equal to
(i) enthalpy of fusion + enthalpy of vapourisation
Which of the following is not correct?
(ii) ∆G is positive for a spontaneous reaction
Thermodynamics mainly deals with
(i) interrelation of various forms of energy and their transformation from one form to another.
(iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.
In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. For such system
(i) qp will be negative
(ii) ∆rH will be negative
The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are
(iii) gas expanding to fill the available volume.
(iv) burning carbon in oxygen to give carbon dioxide.
For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression w = – nRT ln 71 Thermodynamics A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.
(iii) Work done at 600 K is twice the work done at 300 K.
(iv) ∆U = 0 in both cases.
Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below : –1 2 Zn (s) + O2 (g) → 2 ZnO (s) ; ∆H = – 693.8 kJ mol
(i) The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.
(iii) 693.8 kJ mol energy is evolved in the reaction. –1
18.0 g of water completely vapourises at 100°C and 1 bar pressure and the –1 enthalpy change in the process is 40.79 kJ mol . What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?
+ 81.58 kJ, ∆vapH = + 40.79 kJ mol
One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?
Water
Standard molar enthalpy of formation, ∆f H is just a special case of enthalpy V V V of reaction, ∆r H . Is the ∆r H for the following reaction same as ∆f H ? Give reason for your answer. V –1 CaO(s) + CO2(g) → CaCO3(s); ∆f H = –178.3 kJ mol V –1
No, since CaCO3 has been formed from other compounds and not from its constituent elements. V –1
The value of ∆f H for NH3 is – 91.8 kJ mol . Calculate enthalpy change for the following reaction : 2NH3(g) → N2(g) + 3H2(g)
∆r H = +91.8 kJ mol
Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ∆r H and ∆r H1, ∆rH2, ∆r H3 ..... represent enthalpies of intermediate reactions leading to product B. What will be the relation between ∆r H for overall reaction and ∆r H1 , ∆r H2 ..... etc. for intermediate reactions.
∆r H = ∆r H1+ ∆r H2 + ∆r H3 ..... 1665 –1 –1
The enthalpy of atomisation for the reaction CH4(g)→ C(g) + 4H (g) is 1665 kJ mol–1. What is the bond energy of C–H bond?
kJ mol = 416.2 kJ mol –1
Use the following data to calculate ∆lattice H for NaBr. V –1 ∆sub H for sodium metal = 108.4 kJ mol –1 Ionization enthalpy of sodium = 496 kJ mol –1 Electron gain enthalpy of bromine = – 325 kJ mol Bond dissociation enthalpy of bromine = 192 kJ mol–1 V –1 ∆f H for NaBr (s) = – 360.1 kJ mol
+735.5 kJ mol
Given that ∆H = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
It is spontaneous process. Although enthalpy change is zero but randomness or disorder (i.e., ∆ S ) increases. Therefore, in equation ∆G = ∆H – T∆S, the term T∆S will be negative. Hence, ∆G will be negative. qrev
Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.
∆S =
Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?
Yes
At 298 K. Kp for the reaction N2O4 (g) 2NO2 (g) is 0.98. Predict whether the reaction is spontaneous or not.
The reaction is spontaneous ∆r G = – RT ln Kp
A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in Fig. 6.1. What will be the value of ∆H for the cycle as a whole? Fig. : 6.1
This question refers to a figure in the original PDF.
∆H (cycle) = 0
The standard molar entropy of H2O (l ) is 70 J K–1 mol–1. Will the standard molar entropy of H2O(s) be more, or less than 70 J K–1 mol–1?
Less, because ice is more ordered than H2O (l).
Identify the state functions and path functions out of the following : enthalpy, entropy, heat, temperature, work, free energy.
State Functions : Enthalpy, Entropy, Temperature, Free energy Path Functions : Heat, Work 79 Thermodynamics
The molar enthalpy of vapourisation of acetone is less than that of water. Why?
Because of strong hydrogen bonding in water, its enthalpy of vapourisation is more.
Which quantity out of ∆rG and ∆rGV will be zero at equilibrium?
∆rG will always be zero. V V V ∆rG is zero for K = 1 because ∆G = – RT lnK, ∆G will be non zero for other values of K.
Predict the change in internal energy for an isolated system at constant volume. 73 Thermodynamics
For isolated system, there is no transfer of energy as heat or as work i.e., w=0 and q=0. According to the first law of thermodynamics. ∆U = q + w = 0+0=0 ∴ ∆U = 0
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.
At constant volume By first law of thermodynamics: q = ∆U + (–w) (–w) = p∆V ∴ q = ∆U + p∆V ∆V = 0, since volume is constant. ∴ qV = ∆U + 0 ⇒ qV = ∆U = change in internal energy At constant pressure qp = ∆U + p∆V But, ∆U + p∆V = ∆H ∴ qp = ∆H = change in enthalpy. So, at a constant volume and at constant pressure heat change is a state function because it is equal to change in internal energy and change in enthalpy respectively which are state functions.
Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?
(–w) = pext (V2–V1) = 0 × (5 – 1) = 0 For isothermal expansion q = 0 By first law of thermodynamics q = ∆U + (–w) ⇒ 0 = ∆U + 0 so ∆U = 0
Heat capacity (Cp ) is an extensive property but specific heat (c) is an intensive property. What will be the relation between Cp and c for 1 mol of water?
For water, heat capacity = 18 × specific heat or Cp = 18 × c –1 –1 Specific heat = c = 4.18 Jg K –1 –1 Heat capacity = Cp = 18 × 4.18 JK = 75.3 J K
The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.
CP – CV = nR = 10 × 4.184 J
If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
Molar enthalpy = enthalpy change for 1 g carbon × molar mass of carbon change of graphite –1 –1 = – 20.7 kJ g × 12g mol ∴ ∆H = – 2.48 × 102 kJ mol–1 Negative value of ∆H ⇒ exothermic reaction.
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. H2(g) + Br2(g) → 2HBr(g) –1 –1 Given that Bond energy of H2, Br2 and HBr is 435 kJ mol , 192 kJ mol and –1 368 kJ mol respectively. –1
∆rH = Bond energy of H2 + Bond energy of Br2 – 2 × Bond energy of HBr –1 = 435 + 192 – (2 × 368) kJ mol ⇒ ∆rH = –109 kJ mol–1
The enthalpy of vapourisation of CCl4 is 30.5 kJ mol . Calculate the heat required for the vapourisation of 284 g of CCl4 at constant pressure. (Molar –1 mass of CCl4 = 154 g mol ).
qp = ∆H = 30.5 kJ mol–1 284g ∴ Heat required for vapourisation of 284 g of CCl4 = –1 × 30.5 kJ mol–1 154g mol = 56.2 kJ
The enthalpy of reaction for the reaction : –1 2H2(g) + O2(g) → 2H2O(l) is ∆rHV = – 572 kJ mol . What will be standard enthalpy of formation of H2O (l ) ?
According to the definition of standard enthalpy of formation, the enthalpy change for the following reaction will be standard enthalpy of formation of H2O (l) H2(g) + O (g) → H2O(l ). 2 2 or the standard enthalpy of formation of H2O(l ) will be half of the enthalpy of the given equation i.e., ∆rH is also halved. –1 V 1 V − 572 kJ mol ∆f H H O(l ) = × ∆r H = = – 286 kJ/mol. 2 2 2
What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, pext in a single step as shown in Fig. 6.2. Explain graphically.
This question refers to a figure in the original PDF.
Work done on an ideal gas can be calculated from p-V graph shown in Fig. 6.6. Work done is equal to the shaded area ABVIVII . Fig. : 6.6 81 Thermodynamics
How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps? Fig. : 6.2
The work done can be calculated with the help of p–V plot. A p–V plot of the work of compression which is carried out by change in pressure in infinite steps, is given in Fig. 6.7. Shaded area represents the work done on the gas. Fig. : 6.7
Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof. 1 1 H (g) + Cl2(g) HCl(g)
(b) ∆rH = –92.32 kJ mol–1 2 2 2 In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Enthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide spontaneity of a reaction from given diagram. Explain.
This question refers to a figure in the original PDF.
No. Enthalpy is one of the contributory factors in deciding spontaneity but it is not the only factor. One must look for contribution of another factor i.e., entropy also, for getting the correct result.
1 . 0 m o l o f a m o n oa t o m i c i d e a l g a s i s expanded from state (1) to state (2) as shown in Fig. 6.4. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.
This question refers to a figure in the original PDF.
It is clear from the figure that the process has been carried out in infinite steps, hence it is isothermal reversible expansion. V2 w = – 2.303nRT log V V2 p1 2 But, p1V1 = p2V2 ⇒ V = p = =2 1 2 1 p1 ∴ w = – 2.303 nRT log p –1 = – 2.303 × 1 mol × 8.314 J mol K–1 × 298 K–1 × log 2 = – 2.303 × 8.314 × 298 × 0.3010 J = –1717.46 J
An ideal gas is allowed to expand against a Fig. : 6.3 constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J) IV. Matching Type Fig. : 6.4 In the following questions more than one correlation is possible between options of both columns.
w = – pex (Vf –Vi ) = –2 × 40 = – 80 L bar = – 8 kJ The negative sign shows that work is done by the system on the surrounding. Work done will be more in the reversible expansion because internal pressure and exernal pressure are almost same at every step. IV. Matching Type
Match the following : A B (i) Adiabatic process
(a) Heat (ii) Isolated system
(b) At constant volume (iii) Isothermal change
(d) No exchange of energy and matter (v) State function (e) No transfer of heat (vi) ∆U = q (f) Constant temperature (vii) Law of conservation of energy (g) Internal energy (viii) Reversible process (h) pext = 0 (ix) Free expansion (i) At constant pressure (x) ∆H = q (j) Infinitely slow process which proceeds through a series of equilibrium states. (xi) Intensive property (k) Entropy (xii) Extensive property (l) Pressure (m) Specific heat 75 Thermodynamics
(c) First law of thermodynamics (iv) Path function
Match the following processes with entropy change: Reaction Entropy change (i) A liquid vapourises
(a) ∆S = 0 (ii) Reaction is non-spontaneous
(b) ∆S = positive at all temperatures and ∆H is positive (iii) Reversible expansion of an
(c) ∆S = negative ideal gas
Match the following parameters with description for spontaneity : ∆ (Parameters) Description V V V ∆ rH ∆ rS ∆ rG (i) + – +
(a) Non-spontaneous at high temperature. (ii) – – + at high T
(c) Non-spontaneous at all temperatures
(b) Spontaneous at all temperatures (iii) – + –
Match the following : (i) Entropy of vapourisation
(a) decreases (ii) K for spontaneous process
(b) is always positive (iii) Crystalline solid state
(d) Tb of ideal gas
(c) lowest entropy ∆H vap (iv) ∆U in adiabatic expansion
Assertion (A): Combustion of all organic compounds is an exothermic reaction.
Reason (R): The enthalpies of all elements in their standard state are zero. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) A is false but R is true.
(ii) Both A and R are true but R is not the correct explanation of A.
Assertion (A): Spontaneous process is an irreversible process and may be reversed by some external agency.
Reason (R): Decrease in enthalpy is a contributory factor for spontaneity. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) A is false but R is true.
(ii) Both A and R are true but R is not the correct explanation of A.
Assertion (A): A liquid crystallises into a solid and is accompanied by decrease in entropy.
Reason (R): In crystals, molecules organise in an ordered manner. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. (iv) A is false but R is true.
(i) Both A and R are true and R is the correct explanation of A.
Derive the relationship between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.
Extensive properties depend on the quantity of matter but intensive properties do not. Explain whether the following properties are extensive or intensive. Mass, internal energy, pressure, heat capacity, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.
Hint : Ratio of two extensive properties is always intensive Extensive = Intensive . Extensive Moles (Extensive) e.g., Mole fraction = = Total number of moles (Extensive) 1 – V
The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of NaCl(s).
• Na (s) + Cl (g) → Na+(g) + Cl (g) ; ∆latticeH 2 2 83 Thermodynamics • Bonn - Haber Cycle • Steps to measure lattice enthalpy from Bonn - Haber cycle • Sublimation of sodium metal (1) Na(s) → Na (g) ; ∆sub H (2) Ionisation of sodium atoms Na(g) → Na+(g) + e–(g) ; ∆tH i.e., ionisation enthalpy (3) Dissociation of chlorine molecule 1 1 V Cl (g) → Cl(g) ; ∆ H i.e., One-half of bond dissociation 2 2 2 bond enthalpy. (4) Cl(g) + e–(g) → Cl–(g) ; ∆egH i.e., electron gain enthalpy. + Na (g) + Cl (g) 1 V V ∆ H ∆egH 2 bond + 1 Na (g) + Cl2(g) – Na+(g) + Cl (g) ∆iH Na(g) + Cl (g) 2 2 V ∆latticeH ∆subH Na(s) + Cl (g) 2 2 ∆f H NaCl(s)
∆G is net energy available to do useful work and is thus a measure of “free energy”. Show mathematically that ∆G is a measure of free energy. Find the unit of ∆G. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?
∆STotal = ∆Ssys + ∆Ssurr -∆H sys ∆STotal = ∆Ssys + T T ∆STotal = T ∆Ssys – ∆Hsys For spontaneous change, ∆Stotal > 0 ∴ T ∆ Ssys – ∆Hsys > 0 ⇒ – (∆Hsys – T ∆ Ssys ) > 0 But, ∆Hsys – T ∆ Ssys = ∆Gsys ∴ – ∆Gsys > 0 ⇒ ∆Gsys = ∆Hsys – T ∆Ssys < 0 ∆Hsys= Enthalpy change of a reaction. T ∆Ssys = Energy which is not available to do useful work. ∆Gsys = Energy available for doing useful work. • Unit of ∆G is Joule • The reaction will be spontaneous at high temperature.
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 . 77 Thermodynamics
Fig. : 6.9 (i) Reversible Work is represented by the combined areas and . (ii) Work against constant pressure, pf is represented by the area Work (i) > Work (ii) 85 Thermodynamics