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9 months ago

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P
Payal Gupta

Contributor-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+? nRT
                                         ? H = (- 742.7 KJ) + (1/2 mol) x (8.314 x10-3 KJ-1 m

New answer posted

9 months ago

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Payal Gupta

Contributor-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.

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9 months ago

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P
Payal Gupta

Contributor-Level 10

6.  (iv) Possible at any temperature

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Payal Gupta

Contributor-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

...more

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Payal Gupta

Contributor-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.

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New question posted

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New answer posted

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Payal Gupta

Contributor-Level 10

1. (ii) whose value is independent of path

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alok kumar singh

Contributor-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.

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