Thermodynamics

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Payal Gupta

Contributor-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 

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Payal Gupta

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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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Payal Gupta

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11. The combustion equation is:
                                    C (s) + O (g) → CO (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 CO= (393.5 KJ) x (35.2g)/ (44g)

= 314.8 kJ

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Payal Gupta

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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

       

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Payal Gupta

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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.

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Payal Gupta

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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

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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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Payal Gupta

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6.  (iv) Possible at any temperature

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Payal Gupta

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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

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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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