Could you please help me with this problem? thank you!!

#K_(t ot) -= K_c = K_p = 0.635#

#chi_B = 0.285#

Did you try writing out each #K# expression?

#A(g) + B(g) rightleftharpoons C(g)#, #K_1 = 3.36#

#K_1 = ([C])/([A][B])#

#2A(g) + D(g) rightleftharpoons C(g)#, #K_2 = 7.17#

#K_2 = ([C])/([A]^2[D])#

What about the net reaction?

#C(g) + D(g) rightleftharpoons 2B(g)#

#K_(t ot) -= K_c = ([B]^2)/([C][D])#

Here's a wild thought... what if we did this?

#1/K_1^2 = ([A][B])^2/([C])^2 = ([A]^2[B]^2)/([C]^2)#

which would be for the reaction:

#2C(g) rightleftharpoons 2A(g) + 2B(g)#

Starting to look close... What if we did this?

#K_2/K_1^2 = cancel([C])/(cancel([A]^2)[D])(cancel([A]^2)[B]^2)/([C]^cancel(2)) = ([B]^2)/([C][D])#

Does that look familiar? It should.

#color(blue)(K -= K_c = K_2/K_1^2)#

As it turns out, we have just discovered that

• Reversing a reaction flips #K#...
• Multiplying the whole reaction by a constant raises its #K# to that constant.

Therefore...

#color(blue)(K) = 7.17/(3.36)^2 = color(blue)(0.635)#

And if we wish, we can rewrite this in terms of partial pressures by assuming ideality...

#[A] = n_A/V = P_A/(RT)#

Therefore:

#K_c = ([B]^2)/([C][D]) = (n_B/V)^2/(n_C/V n_D/V)#

#= (P_B/(RT))^2/(P_C/(RT) P_D/(RT))#

#= P_B^2/(P_CP_D) cdot cancel(1/(RT)^2 cdot (RT)(RT))#

But did we really change anything? No... The #RT# all cancel out, and the volumes all cancel out, so

#([B]^2)/([C][D]) = n_B^2/(n_Cn_D) = P_B^2/(P_CP_D)#.

The change in mols of gas is zero, so NUMERICALLY, #color(blue)(K_c = K_p)#...

#=> color(blue)(K_p = 0.635)# in terms of #"atm"#

#= (P_B^2)/("1.53 atm" - x)^2#

And again, this is for the reaction:

#C(g) + D(g) rightleftharpoons 2B(g)#

Here we have a coefficient of #2# for #B#, so #P_B -= 2x# and #P_B^2 = 4x^2#. Therefore:

#0.635 = (4x^2)/(1.53 - x)^2#

And since this is a perfect square,

#sqrt(0.635) = (2x)/(1.53 - x)#

(We choose #+sqrt(0.635)# so that #K > 0# for physical reasons.)

#sqrt(0.635)(1.53 - x) = 2x#

#sqrt(0.635)cdot1.53 = 2x + sqrt(0.635)x#

As a result:

#color(blue)x = (sqrt(0.635)cdot1.53)/(2 + sqrt(0.635)) = color(blue)"0.436 atm"#

Therefore, the total pressure at constant temperature is:

#P_(t ot) = P_B + P_C + P_D#

#= 2x + (1.53 - x) + (1.53 - x)#

#= cancel(2x) + 2(1.53) - cancel(2x)#

#=# #"3.06 atm"#

And since

#P_B = chi_BP_(t ot)#,

the mol fraction of #B# is:

#color(blue)(chi_B) = P_B/P_(t o t)#

#= (2x)/(3.06) = 0.872/3.06 = color(blue)(0.285)#