Thermodynamics
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Assertion and Reason:
Directions:
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
23. Assertion: Variables like pressure (p), volume (V), temperature (T) etc. are called state variables or state functions.
Reason: Their values depend only on the state of the system and not on how it is reached
Assertion and Reason:
Directions:
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
23. Assertion: Variables like pressure (p), volume (V), temperature (T) etc. are called state variables or state functions.
Reason: Their values depend only on the state of the system and not on how it is reached
23. (a) The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T), amount (n) etc. Variables like p, V, T are called state variables or state functions because thei
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22. Calculate the entropy change in surroundings when 1.00 mol of H2O (l) is formed under standard conditions. ΔfH? = – 286 kJ mol–1.
22. Calculate the entropy change in surroundings when 1.00 mol of H2O (l) is formed under standard conditions. ΔfH? = – 286 kJ mol–1.
22. qrev = (-ΔfH? ) = - (- 286kJ mol-1) = 286 x 103 Jmol-1 = 286000 J mol-1
ΔS (surroundings) = qrev/ T = 286000 J mol-1 / 298 K
Upvote 21. Comment on the thermodynamic stability of NO (g), given
½ N2 (g) + ½ O2 (g) → NO (g); ΔrH? = 90 kJ mol–1
NO (g) + ½ O2 (g) → NO2 (g); ΔrH? = –74 kJ mol–1
21. Comment on the thermodynamic stability of NO (g), given
½ N2 (g) + ½ O2 (g) → NO (g); ΔrH? = 90 kJ mol–1
NO (g) + ½ O2 (g) → NO2 (g); ΔrH? = –74 kJ mol–1
21. For NO (g), ΔrH° is a +ve value. So, it is unstable in nature.
For NO2 (g), ΔrH° is a -ve value. So, it is stable
Upvote 20. The equilibrium constant for a reaction is 10. What will be the value of ΔG°?
R = 8.314 JK–1 mol–1, T = 300 K.
20. The equilibrium constant for a reaction is 10. What will be the value of ΔG°?
R = 8.314 JK–1 mol–1, T = 300 K.
20. ΔG° = -RT ln K = - 2.303 RT log K
Putting the values of
R = 8.314 J K-1 mol-1,
T = 300 K and
K =10; we get
ΔG° = - 2.303 x 8.314 J K-1 mol-1 x 300 K and K x log 10
= - 5527 J mol-1
= -5.527 kJ mol
Upvote 19. For the reaction 2 A(g) + B(g) → 2D(g)
ΔU° = – 10.5 kJ and ΔS? = – 44.1 JK–1.
Calculate ΔG? for the reaction, and predict whether the reaction may occur spontaneously.
19. For the reaction 2 A(g) + B(g) → 2D(g)
ΔU° = – 10.5 kJ and ΔS? = – 44.1 JK–1.
Calculate ΔG? for the reaction, and predict whether the reaction may occur spontaneously.
19. ΔH° = ΔU° + Δng RT
ΔU° = -10.5 kJ, Δng = 2-3 = -1 mol, R = 8.314 x 10-3 kJ mol-1, T = 298 K
ΔH° = (- 10.5 kJ) + [ (- 1 mol) x (8.314 x 10-3 kJ mol-1) x (298 K)]
= -10.5 kJ – 2.478 kJ
= -12.978 kJ
According to Gibbs Helmhol
Upvote 18. For the reaction; 2Cl (g) → Cl2 (g); what will be the signs of ?H and ?S?
18. For the reaction; 2Cl (g) → Cl2 (g); what will be the signs of ?H and ?S?
18. ? H: negative (-ve) because energy is released in bond formation
? S: negative (-ve) because entropy decreases when atoms combine to form molecules.
Upvote 17. For the reaction at 298 K, 2A + B → C
ΔH = 400 kJ mol–1 and ΔS = 0.2 kJ K–1 mol–1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
17. For the reaction at 298 K, 2A + B → C
ΔH = 400 kJ mol–1 and ΔS = 0.2 kJ K–1 mol–1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
17. As per the Gibbs Helmholtz’s equation:
ΔG = Δ H - TΔ S
For ΔG=0;
ΔH=TΔS
Or T=ΔH/ΔS
T = (400 KJ mol-1)/ (0.2 KJ K-1 mol-1)
= 2000 k
Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature
Upvote 16. Change in internal energy (? U) for an isolated system is zero for it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore? S > 0 or positive.
Upvote 15. Calculate the enthalpy change for the process
CCl4(g) → C(g) + 4 Cl(g)
and calculate bond enthalpy of C – Cl in CCl4(g).
ΔvapH?(CCl4) = 30.5 kJ mol–1.
ΔfH? (CCl4) = –135.5 kJ mol–1.
ΔaH? (C) = 715.0 kJ mol–1, where ΔaH? is enthalpy of atomisation
ΔaH? (Cl2) = 242 kJ mol–1
15. Calculate the enthalpy change for the process
CCl4(g) → C(g) + 4 Cl(g)
and calculate bond enthalpy of C – Cl in CCl4(g).
ΔvapH?(CCl4) = 30.5 kJ mol–1.
ΔfH? (CCl4) = –135.5 kJ mol–1.
ΔaH? (C) = 715.0 kJ mol–1, where ΔaH? is enthalpy of atomisation
ΔaH? (Cl2) = 242 kJ mol–1
15. According to the question:
i. CCl4 (l) → CCl4 (g), ? vapH? = 30.5 kJ mol–1
iii. C (s) + 2Cl2 (g) → CCl4 (l), ? fH? = –135.5 kJ m
Upvote 14. Calculate the standard enthalpy of formation of CH3OH from the following data:
(i) CH3OH(l) + 3/2 O2 (g) → CO2 (g) + 2H2O (l); ?rH?= – 726kJ mol-1
(ii) C(graphite) + O2(g) → CO2 (g); ?cH? = – 393 kJ mol-1
(iii) H2(g) + 1/2O2(g) → H2O (l); ?fH? = -286 kJ mol-1
14. Calculate the standard enthalpy of formation of CH3OH from the following data:
(i) CH3OH(l) + 3/2 O2 (g) → CO2 (g) + 2H2O (l); ?rH?= – 726kJ mol-1
(ii) C(graphite) + O2(g) → CO2 (g); ?cH? = – 393 kJ mol-1
(iii) H2(g) + 1/2O2(g) → H2O (l); ?fH? = -286 kJ mol-1
14. The reaction for the formation of CH3OH can be obtained by:
C (s) + 2H2 (g) + l/2O2 (g) → CH3OH (l)
This can be obtained by the algebraic calculations of the reactions:
Equation (ii) + 2 x equation (iii) – equation (i)
? Hf [CH3OH] =? cH? + 2 x? fH? -? rH?
&n
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