Ncert Solutions Chemistry Class 11th
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New answer posted
5 months agoContributor-Level 10
11. The combustion equation is:
C (s) + O2 (g) → CO2 (g); ? H = – 393.5 KJ mol-1
Heat released in the formation of 44g of CO2 = 393.5 kJ
Heat released in the formation of 35.2 g of CO2 = (393.5 KJ) x (35.2g)/ (44g)
= 314.8 kJ
New answer posted
5 months agoContributor-Level 10
10. Total enthalpy change involved in the transformation is the sum of the following changes:
(i) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C
ΔH= Cp [H2O (l)] ΔT
(ii) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of ice at 0°C
ΔHfreezing
(iii) Energy change involved in the transformation of 1 mol of ice at 10°C to 1 mol of ice at 0°C
 
New answer posted
5 months agoContributor-Level 10
9. No. of moles of Al, n = (60g)/ (27 g mol-1) = 2.22 mol
Molar heat capacity (C) = 24 J mol-1 K-1.
Rise in temperature (? T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C x m x T = (24 J mol-1 K-1) x (2.22 mol) x (20 K) = 1065.6 J = 1.067 KJ.
New answer posted
5 months agoContributor-Level 10
8. Given,
? U = – 742.7 KJ-1 mol-1
? ng = 2 – 3/2 = + 1/2 mol.
R = 8.314 x 10-3KJ-1 mol-1
T = 298 K
According to the relation? H =? U+? ng RT
? H = (- 742.7 KJ) + (1/2 mol) x (8.314 x10-3 KJ-1 m
New answer posted
5 months agoContributor-Level 10
7. As per the first law of thermodynamics,
? U= q + W
Heat absorbed by the system, q = 701 J
Work done by the system, W = – 394 J
Change in internal energy? U = q + w = 701 – 394 = 307 J.
New answer posted
5 months agoContributor-Level 10
5. As per question:
(i) CH4? +2O2? CO2? +2H2? O ΔH1? =−890kJmol−1
(ii) C+O2? CO2? ΔH2? =−393.5kJmol−1
(iii) 2H2? +O2?2H2? O ΔH3? =2* (−285.8)kJmol−1
Required reaction is
C+2H2? CH4 (g)? ; ΔHf? =?
From equat
New answer posted
5 months agoContributor-Level 10
4.The chemical equation for the combustion reaction is:
CH4 (g) + 2O2 (g) →CO2 (g) + 2H2O (l)
Δng= 1 – 3 = -2
ΔH? = ΔU? + ΔngRT = ΔU? - 2RT
Therefore, ΔH? <ΔU?
i.e. option (iii) is correct.
New answer posted
5 months agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
At constant volume
q = ΔU + (-w)
-w = pΔ q = AU + pΔV
ΔV = 0 (at constant volume)
Hence, qv = ΔU + 0 = ΔU= change in internal energy at constant pressure, qp = AU + pΔV
Since ΔU + pΔV=ΔH
=> qp = ΔH change in enthalpy
Hence, at constant volume and at constant pressure, heat change is a state function because it is equal to ΔU and ΔH respectively which are state functions.
New answer posted
5 months agoContributor-Level 10
This is a Short Answer Type Questions as classified in NCERT Exemplar
For an isolated system w = 0, q = 0
Since ΔU= q + w = 0 + 0 = 0, ΔU= 0
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