Question

When heated at high temperatures, a diatomic vapor disassociates as follows:

A₂(g) ⇌ 2A(g)

In one experiment, a chemist finds that when 0.0490 mole of A₂ was placed in a flask of volume 0.478 L at 584 K, the degree of dissociation (that is, the fraction of A₂ dissociated) was 0.0246. Calculate K_c and K_P for the reaction at this temperature.

Kc= 0.000252

I NEED THE ANSWER TO Kp!!!!!!!!!!!

Answer 1

Determine the relation between K_c and K_p for the equilibrium reaction as below.

K_c = [A]²/[A₂]

Where the square braces denote molar concentrations.

The partial pressures for the two gaseous components can be written as

P_A2V = n_A2RT

Where V is the volume of the container and T is the absolute temperature of the reaction; n_A2 is the number of moles of A₂ in the reaction container.

Therefore,

P_A2 = (n_A2/V)*RT

= [A₂]RT

Where [A₂] = n_A2/V is the molar concentration of A₂ in the reaction flask.

Similarly, P_A = [A]RT

The equilibrium constant K_p can be written as

K_p = (P_A)²/(P_A2)

= {[A]RT}²/{[A₂]RT}

= [A]²/[A₂]*RT

= K_c*(RT)

The value of K_c is already known. Use the known values of K_c, R and T (all without units since equilibrium constants are dimensionless) to determine K_p as

K_p = (0.000252)*(0.082 L-atm/mol.K)*(584 K)

= 0.01206

≈ 0.0121 (ans, correct to 3 sig. figs).