Asystem was prepared with nh3 = o2 = 3.60 m as the only components initially. at equilibrium n2o4 is 0.60m. calculate the value of the equilibrium constant kc for the reaction

Kc = 0.0216

Explanation:

To do this, we need to write the equilibrium reaction which is the following:

NH3 + O2 <> N2O4 + H2O

And now, let's balance the equation

4NH3 + 7O2 <> 2N2O4 + 6H2O

The general expression for Kc is:

Kc = [N2O4]²[H2O]⁶ / [O2]⁷[NH3]⁴

Now, in order to know the concentrations, we need to do an ICE chart here, so we can know the final concentrations of all species here:

         4NH3 + 7O2 <> 2N2O4 + 6H2O

I:          3.6        3.6                      0            0

C:         -4x       -7x                      +2x        +6x

E:      3.6 - 4x   3.6 - 7x              0.6          +6x

According to this, we know that N2O4 is 0.6M, but in the change we have that it gained 2x so the value of x will be:

2x = 0.6

x = 0.6/2 = 0.3

Therefore, the value of x is 0.3 ans we can know now the concentrations of all species in equilibrium, and then, the kc:

[NH3] = 3.6 - 4(0.3) = 2.4 M

[O2] = 3.6 - 7(0.3) = 1.5 M

[H2O] = 6(0.3) = 1.8 M

Now the value of Kc would be:

Kc = (0.6)² (1.8)⁶ / (2.4)⁴ (1.5)⁷

Kc = 0.0216

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.