Kc for reaction 2 at the same temperature
[tex]\large{\boxed{\bold{Kc~=~50^2~=~2500}}}[/tex]
[tex]\boxed{\boxed{\bold{Further~explanation}}}[/tex]
The equilibrium constant or Kc is the price of the concentration product in the equilibrium state of the substance in the right segment divided by the product of the substance in the left section, each of which has a reaction coefficient raised
The equilibrium constant based on concentration (Kc) in a reaction
pA + qB -----> mC + nD
[tex]\large{\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}[/tex]
[tex]Unit~Kc~is~M ^{(p + q) - (m + n)}[/tex]
In reaction 1 2CH₄ (g) ⇄ C₂H₂ (g) + 3H₂ (g) Kc = 0.020
On reaction 2: 2C₂H₂ (g) + 6H₂ (g) ⇄ 4CH₄ (g)
From the 2 equation the above reaction shows that reaction 2 is the opposite of reaction 1 whose coefficient is multiplied by 2. Then the price of Kc reaction 2 is the square of the inverse price of Kc reaction 1
2CH₄ (g) ⇄ C₂H₂ (g) + 3H₂ (g) Kc = 0.020
[tex]\large{\boxed{\bold{Kc~=~\frac{[C_2H_2][H_2]^3}{[CH_4]^2} }}}[/tex]
Then reversed
C₂H₂ (g) + 3H₂ (g) ⇄ 2CH₄ (g)
[tex]\large{\boxed{\bold{Kc~=~\frac{[CH_4]^2}{[[C_2H_2][H_2]^3} }}}[/tex]
Kc = 1 / 0.02 = 50
Then reaction 2 :
2C₂H₂ (g) + 6H₂ (g) ⇄ 4CH₄ (g)
[tex]\large{\boxed{\bold{Kc~=~\frac{[CH_4]^4}{[[C_2H_2]^2[H_2]^6} }}}[/tex]
[tex]\large{\boxed{\bold{Kc~=~50^2~=~2500}}}[/tex]
[tex]\boxed{\boxed{\bold{Learn~more}}}[/tex]
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Keywords: constant equilibrium, Kc, concentration, product, reactant, reaction coefficient
