Question

I NEED A COMPLETE AND CLEAR SOLUTION PLEASE! thanks

A mixture of 0.10 mol of NO, 0.050 mol of H2 and 0.10 mol of H2O is placed in a 1.0-L flask and allowed to reach equilibrium as shown below: 2 NO (g) +2 H(g)N2 (g) + 2 H2O (g) At equilibrium [NO] = 0.062 M. Calculate the equilibrium constants Kc and Kp for this reaction.

Answer 1

Initially

[NO] = 0.1/1 = 0.1 M

[H2] = 0.05/1 = 0.05 M

[N2] = 0

[H2O] = 0.1/1 = 0.1 M

At equilibrium

[NO] = 0.062 M

[NO ] decreased by 0.10-0.062 = 0.038 M

[H2] = 0.05 -0.038 = 0.012 M

[N2] = 0.038 M

[H2O] = 0.1 + 0.038 = 0.138 M

Kc = [N2][H2O]²/[NO]²[H2]²

   = 0.038 x (0.138)²/(0.062)²(0.012)² = 0.000723672/0.00000055353 = 1307.37

del n = n_p - n_r = 3-4 = -1

Kp = Kc(RT)^{del n}

= Kc(0.082 x 298)^-1

= 1307.37(24.436)^-1 = 53.50