Explanation
For an ideal gas, the heat of reaction at constant pressure and constant volume are related as qp = qv + ∆nRT
Given, 1) C(grahite) + O2(g) → CO2 (g) ; ∆H = -393.5 KJ
2) H2(g) + 1/2 O2 (g) → H2O(l); ∆H = -286.2KJ
3) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O; ∆H = -1410.8KJ
Now, formation of ethylene molecule.
2C + 2H2 →C2H4
∴ Multiply equation 1 & I2 by 2, we get,
4) 2C + 2O2 → 2CO2 ; ∆H = -787.0KJ
5) 2H2 + O2 → 2H2O; ∆H = -572.4KJ
Add (4) and (5),
4) 2C+2H2 +3O2→2CO2+2H2O; ∆H = -1359.4KJ
Subtract (5) from (3)
2C + 2H2 →C2H5; ∆H = 51.4KJ
In thermodynamics, the Joule-Thomson effect or Joule-Thomson expansion describes the temperature change of a real gas or liquid (as differentiated from an ideal gas) when it is forced through a value or porous plug while kept insulted so that no heat is exchanged with the environment. This procedure is called a throtting process or joule-Thomson process . At room temperature, all gases except hydrogen, helium and neon cool upon expansion by the joule- Thomson process , these three gases experience the same effect but only at lower temperatures.
The absolute value of enthalpy (H) cannot be determined experimentally because as many of the contributing forms cannot be known. Whereas change in enthalpy ΔH can be determined experimentally.
Total energy of the universe remains constant.
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