Answer to Question #153528 in General Chemistry for Buddika

Question

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

Answer 1

Solution:

1) Given chemical reaction:

A₂B₄(g) - 2AB_2(g)

$\def\arraystretch{1.5} \begin{array}{c:c:c} I & 0.85 & 0 \\ \hline C & -x & 2x \\ \hdashline E & 0.98 & -0.26 \end{array}$

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.

$K_p=K_c(RT)^{n(gas)}$

$n_{gas}=2-1=1$

$K_c=\tfrac{K_p}{RT}=\tfrac{0.069}{0.0821*273}$

$K_c=0.003$