Question

The equilibrium constant Kc for the reaction,
A(g) + 2B(g) -----> 3C(g) is 2 x 10^-3
What would be the equilibrium partial pressure of gas C if initial pressure of gas A & B are 1 & 2 atm respectively
(a) 0.0625 atm
(b) 0.1875 atm
(C) 0.21 atm
(d) None of these​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

(b)0.1875 atm

Step-by-step explanation:

$K_c=2\times 10^{-3}$

Number of moles of reactants=1+2=3

Number of moles of product=3

Change in number of moles=3-3=0

$\Delta n=0$

The relation between Kp and Kc

$K_p=K_c(RT)^{\Delta n }$

Substitute the values

$K_p=K_c(RT)^0=K_c=2\times 10^{-3}$

Initial  pressure of gas A=1 atm

Initial pressure of gas B=2 atm

At equilibrium

Partial pressure of gas A=1-x

Partial pressure of gas B=2(1-x)

Partial pressure of gas C=3x

We know that

$K_p=\frac{(P_C)^c}{(P_A)^a(P_B)^b}$

Where $P_A$,$P_B$ =Partial pressure of gas A and gas B

$P_C=$Partial pressure of gas C

Using the formula

$2\times 10^{-3}=\frac{(3x)^3}{(1-x)(2-2x)^2}$

$2\times 10^{-3}=\frac{27x^3}{(1-x)\times 2^2(1-x)^2}$

$\frac{2\times 10^{-3}\times 4}{27}=\frac{x^3}{(1-x)^3}=(\frac{x}{1-x})^3$

$\frac{8}{10^3\times 27}=(\frac{x}{1-x})^3$

By using property $x^{-a}=\frac{1}{x^a}$

We know that

$10^3=1000$

Therefore, $\frac{8}{1000\times 3^3}=(\frac{x}{1-x})^3$

$\frac{2^3}{(30)^3}=(\frac{x}{1-x})^3$

$(\frac{2}{30})^3=(\frac{x}{1-x})^3$

$\frac{x}{1-x}=\frac{2}{30}=\frac{1}{15}$

When power on both sides are equal then base will be equal.

$15x=1-x$

$15x+x=1$

$16x=1$

$x=\frac{1}{16}$

Substitute the value of x

The partial pressure of gas C=$3\times \frac{1}{16}=0.1875 atm$

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