Ncert Solutions Chemistry Class 11th
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New answer posted
5 months agoContributor-Level 10
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
New answer posted
5 months agoContributor-Level 10
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-1
New answer posted
5 months agoContributor-Level 10
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 Helmholtz equation:
ΔG° = ΔH° - TΔS°
= (- 12.978 kJ) – (298 K) x (- 0.0441 kJ K-1)
= -12.978 + 13.142
= 0.164 kJ
Since the value of ΔG° is positive, the reaction is non-spontaneous.
New answer posted
5 months agoContributor-Level 10
18. ? H: negative (-ve) because energy is released in bond formation
? S: negative (-ve) because entropy decreases when atoms combine to form molecules.
New answer posted
5 months agoContributor-Level 10
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
New answer posted
5 months agoContributor-Level 10
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.
New answer posted
5 months agoContributor-Level 10
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 mol–1
iii. C (s) → C (g), ? aH? = 715.0 kJ mol–1
iv Cl2 (g) → C (g) + 4 Cl (g) ? aH? = 242 kJ mol–1
The equation we want is:
CCl4
New answer posted
5 months agoContributor-Level 10
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?
= (– 393 kJ mol-1) + 2 (- 286 kJ mol-1) – (– 726 kJ mol-1)
= (– 393 – 572 + 726) kJ
New answer posted
5 months agoContributor-Level 10
13. N2 (g) + 3H2 (g) → 2NH3 (g); ? rH= -92.4 kJ/mol
Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of a substance in its standard form from its constituent elements in their standard state.
Re-writing the given equation for 1 mol of NH3
1/2N2 (g) + 3/2H2 (g) → NH3
New answer posted
5 months agoContributor-Level 10
12. Given? fH of CO (g) = – 110 kJ mol-1
? fH of CO2 (g) = – 393 kJ mol-1
? fH of N2O (g) = 81 kJ mot-1
? fH of N2O4 (g) = 9.7 kJ mol-1
Enthalpy of reaction (? rH) = [81 + 3 (- 393)] – [9.7 + 3 (- 110)]
= [81 – 1179] – [9.7 – 330]
= – 778 kJ mol-1
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