Answer in General Chemistry for Buddika #153528

A₂B₄(g) – 2AB_2(g) initial pressure of A2B4 in closed system was 0.85 atm and the final equilibrium pressure became 0.98 atm .so calculate Kp and Kc

Solution:

1) Given chemical reaction:

A₂B₄(g) – 2AB_2(g)

“defarraystretch{1.5}n begin{array}{c:c:c}n I & 0.85 & 0 \ hlinen C & -x & 2x \n hdashlinen E & 0.98 & -0.26nend{array}”

0.98-(-x)=0.85

x=- 0.13

“K_p=tfrac{p[AB_2]^2}{p[A_2B_4]}=tfrac{0.26^2}{0.98}”

“K_p=0.069”

2) Concentration constant:

In this case, the temperature is not given in this question, so we consider that we are in standard condition.

“K_p=K_c(RT)^{n(gas)}”

“n_{gas}=2-1=1”

“K_c=tfrac{K_p}{RT}=tfrac{0.069}{0.0821*273}”

“K_c=0.003”