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Published on: 09/09/2022
QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Chemistry Subject - Thermodynamics, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
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1.
We can measure the transfer of heat from one system to another which cause change in temperature. The magnitude of change in temperature depends upon heat capacity of the substance. The enthalpy change of reaction remains the same irrespective of number ofsteps is Hess's law.It helps to calculate enthalpy of formation, combustion and other enthalpy changes. Enthalpy change can also be calculated by using bond enthalpies. First law gives law of conservation of energy but does not give direction of reaction. Second law states, entropy of universe is continuously increasing due to spontaneous processes taking place in it. \(\Delta\)H and \(\Delta\)S (entropy change) cannot decide spontaneity of process. We need \(\Delta\)G (free energy change) which is -ve for spontaneous, +ve for non-spontaneous. \(\Delta\)G = 0 for process in equilibrium. \(\Delta\)G is related to equilibrium constant. If \(\Delta\)G = -ve, 'K' is +ve and vice versa. Third law of thermodynamics states the entropy of perfectly crystalline substance is zero at zero kelvin.
(a) We can determine \(\Delta\)H lattice with the help of cycle. Name the cycle.
(b) How can we calculate enthalpy of solution?
(c) What is molar heat capacity of water in equilibrium with ice at constant pressure?
(d) \(\Delta \mathbf{H}_{f}^{\circ}\) of O3, CaO, H3 and HI are +142.2,- 643.9,-46, +25.95 kJ mol-1. Arrange these in increasing order of stability.
(e) Standard entropy of X2 , Y2 and XY3 are 60, 40 and 50 JK-1 mol-1 respectively. For the reaction \(\frac{1}{2} \mathbf{X}_{2}+\frac{3}{2} \mathbf{Y}_{2} \longrightarrow \mathbf{X Y}_{3}, \Delta \mathbf{H}=-\mathbf{3 0}\) KJ to be at what temperature, process will be at equilibrium.
(f) What are sign of \(\Delta\)H and \(\Delta\)S for process to be always spontaneous?
(g) Give mathematical expression for second law of thermodynamics.
2.
Observe the table of standard enthalpy change for fusion and vapourisation of some substances. Study the table and answer the following questions.
Standard Enthalpy Changes of Fusion and Vapourisation
| Substance | Tf/K | \(\Delta_{f u s} \mathbf{H}^{\odot} /\left(\mathbf{k} \mathbf{J} \mathbf{m o l}^{-1}\right)\) | Tb/K | \(\Delta_{v a p} \mathbf{H}^{\odot} /\left(\mathbf{k} \mathbf{J} \mathbf{m o l}^{-1}\right)\) |
| N2 | 63.15 | 0.72 | 77.35 | 5.59 |
| NH3 | 195.40 | 5.65 | 239.73 | 23.35 |
| HCl | 159.0 | 1.992 | 188.0 | 16.15 |
| CO | 68.0 | 6.836 | 82.0 | 6.04 |
| CH3COCH3 | 177.8 | 5.72 | 329.4 | 29.1 |
| CCl4 | 250.16 | 2.5 | 349.69 | 30.0 |
| H2O | 273.15 | 6.01 | 373.15 | 40.79 |
| NaCI | 1081.0 | 28.8 | 1665.0 | 170.0 |
| C6H6 | 278.65 | 9.83 | 353.25 | 30.8 |
(a) Why is \(\triangle\)Hvap of NaCI highest?
(b) What is \(\triangle\)s for H2O(s) \(\rightarrow\) H2O(I)?
(c) Which has stronger intermolecular forces, CCl4 or Acetone?
(d) calculate \(\triangle\)Uvap for H2O if \(\triangle\)Hvap = 40.66 kJ mol-1 at 373 K for 1 mole.
(e) NH3(g) \(\rightarrow\) NH3 (l). What is sign of \(\triangle\)s?
(f) How is enthalpy of sublimation calculated?
(g) What is relationship between ionisation enthalpy and ionisation energy?
3.
Thermodynamics involve energy changes in chemical reactions and other processes. Internal energy is total energy stored in a substance. We can specify absolute value of volume but not the absolute value of internal energy. We can measure only change in internal energy (\(\triangle\)U). Work done on the system is taken as positive and work done by the system is taken as negative. Heat (q) absorbed by the system is +ve and heat given out by system is negative. \(\triangle\)U = q + w according to first law of thermodynamics. \(\triangle\)H (enthalpy change) is measured at constant pressure, \(\triangle\)U is measured at constant volume. \(\triangle\)H, \(\triangle\)S (entropy change), \(\triangle\)G (free energy change) and temperature help to decide spontaneity of the process.
(a) What is \(\triangle\)U in adiabatic process?
(b) If 701 J of heat is absorbed by the system and 394 J of work is done by the system. What is value of\(\triangle\) U?
(c) 2 litres of an ideal gas at a pressure of 10 atm expands isothermally into vacuum until its total volume is 10 litres. How much heat is absorbed and work done in the expansion?
(d) For an equilibrium H2 O(1) \(\rightleftharpoons\) H2 O(g), What are sign of \(\triangle\)G, \(\triangle\)H and \(\triangle\)S?
(e) For N2 O4 (g) \(\rightleftharpoons\) 2NO2 (g).
(f) What is Cp- Cv equal to?
(g) State second law of thermodynamics.
1.
(a) Born-Haber cycle.
(b) \(\Delta \mathrm{H}_{\text {solution }}^{\circ}=\Delta \mathrm{H}_{\text {lattice }}^{\circ}+\Delta \mathrm{H}_{\text {hydration }^{\circ}}^{\circ}\)
(c) \(\infty\) (infinity)
\(\left.C_{p}=\frac{H_{2}-H_{1}}{\Delta T}=\frac{\Delta H}{0}=\infty \ \text { [At equilibrium } \Delta T=0\right]\)
(d) O3 < HI < NH3 < CaO
(e) \(\Delta S=50-\frac{1}{2} \times 60-\frac{3}{2} \times 40\)
= -40 JK-1 mol-1
\(\Delta S=\frac{\Delta H_{\mathrm{rev}}}{T}\)
\(\Rightarrow \quad T=\frac{-30 \times 1000 \mathrm{~J}}{-40 \mathrm{JK}^{-1}}=750 \mathrm{~K}\)
(f) \(\Delta\)H = -ve, \(\Delta\)S = +ve.
(g) \(\Delta\)S total > 0.
2.
(a) It is due to strong forces of attraction between Na + and Cl-.
(b) \(\Delta S_{\text {fusion }}=\frac{\Delta H_{\text {fusion }}}{\text { Melting point in } K}\)
\(=\frac{6.01 \times 1000 \mathrm{~J}}{273.15 \mathrm{~K}}=22 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)
(c) CCl 4 has more intermolecular forces of attraction..
(d) H2O(l) \(\rightarrow\) H2O(g); \(\triangle\)n = 1
\(\triangle\) H = \(\triangle\)U+ \(\triangle\)nRT
\(\triangle\)U = \(\triangle\)H - \(\triangle\)nRT
= 40.66 kJ - 373 x 8.314 x 10- 3 kJ
\(\triangle\)U = 37.56 kJ mol-1
(e) \(\triangle\)s = -ve.
(f) \(\Delta \mathrm{H}_{\text {sub }}^{\circ}=\Delta \mathrm{H}_{\text {fusion }}^{\circ}+\Delta \mathrm{H}_{\text {vapourisation }^{\circ}}^{\circ}\)
(g) \(\triangle\)rHo (ionisation enthalpy) = Eo (ionisation energy) \(+\frac{5}{2} \mathrm{RT}\)
3.
(a) q = 0, \(\triangle\)U = W adiabatic i.e., internal energy change is equal to work done.
(b) \(\triangle\)U = q - W = 701- 394 = 307 J [ because work is done by the system]
(c) q = - W = Pext (V2 - V1) [\(\because\) Pext = 0 in vacuum]
= 0 x (10 - 2) = 0
No work is done, no heat is absorbed.
(d) \(\triangle\)G = 0, \(\triangle\)H > 0, \(\triangle\)S > 0.
(e) \(\triangle\)H = \(\triangle\)U + \(\triangle\)nRT
\(\triangle\)H =\(\triangle\)U + RT [ \(\because\)\(\triangle\)n = 1]
(f) Cp - Cv = R.
(g) The entropy of universe is continuously increasing due to spontaneous processes taking place.
\(\Delta \mathrm{S}_{\text {sys }}+\Delta \mathrm{S}_{\text {surr }}>0 \text { OR } \ \Delta \mathrm{S}_{\text {Total }}>0\)
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