Question

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=? Kgoal=? by making use of the following information: 1. 2CO2(g)+2H2O(l)⇌CH3COOH(l)+2O2(g)2CO2(g)+2H2O(l)⇌CH3COOH(l)+2O2(g), K1 = K1 = 5.40×10−165.40×10−16 2. 2H2(g)+O2(g)⇌2H2O(l)2H2(g)+O2(g)⇌2H2O(l), K2 = K2 = 1.06×10101.06×1010 3. CH3COOH(l)⇌2C(s)+2H2(g)+O2(g)CH3COOH(l)⇌2C(s)+2H2(g)+O2(g), K3 = K3 = 2.68×10−92.68×10−9

Answer 1

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$.

Explanation:

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$

$K_{goal}=?$

$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] +  [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

 ( on adding the equilibrium constant will get multiplied with each other)

$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

$K=\frac{[C]^2[O_2]^2}{[CO_2]^2}$

On comparing the K and $K_{goal}$:

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$.