All parts in details please One mole of N_2 O_4 is placed in a vessel. When the

Question

All parts in details please

One mole of N_2 O_4 is placed in a vessel. When the equilibrium N_2O_4(g) reversible 2NO_2(g) is established, the enthalpy of the equilibrium mixture is H = (1 - E)H bar ^0(N_2O_4,g) + 2EH bardegree (NO_2,g). What assumption are we making by using molar enthalpies of pure substances here If the mixture remains in equilibrium as the temperature is raised, show that the heat capacity is given by C_p/R = (1 - E) C bar_p(N_2 O_4, g)/R + 2E C bar_p(NO_2,g)/R + 1/2 E(1 - E^2) (delta H^0)/(RT)^2; Show that the last term has a maximum value when E = 1/root 3; Plot C_p/R versus T from 200K to 500K at 1 atm using C bar_p(N_2 O_4,g)/R = 9.29, C bar_P(NO_2, g)/R = 4.47, Delta H^o_298 = 57.20kJ/mol; and delta G^o_298 = 4. HkJ/mol.

Explanation / Answer

Assumption:

1. Enthalpy changes of combustion (Hc ) and formation (Hf ) should be covered.

2. heat change when the temperature of a pure substance is changed using q.

3. The standard state of a solid or liquid substance is the pure element.

N2(g) + 2O2(g) N2O4(l).

The proceeding section shows us a way to measure the heat of a reaction:

Now plot Cp/R versus T,

Substituting the values of subdivision c) and b) in a). we get,

At 200k, Cp/R =28.917

At 300k, Cp/R = 28.911

At 400k, Cp/R = 28.914

At 500k, Cp/R = 28.909

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