Chemistry, 04.08.2019 21:30 cutiecat66

Given the information for reaction 1 at a specific temperature; calculate kc for reaction 2 at the same temperature. reaction 1: 2ch4 (g) ⇄ c2h2 (g) + 3h2 (g) kc = 0.020 reaction 2: 2c2h2 (g) + 6h2 (g) ⇄ 4ch4 (g)

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Answer is: equilibrium constants is 2500.
Chemical reaction 1: 2CH₄(g) ⇄ C₂H₂(g) + 3H₂(g); Kc₁ = 0,020.
Chemical reaction 2: 2C₂H₂(g) + 6H₂(g) ⇄ 4CH₄(g); Kc₂ = ?
Chemical reaction 2 is reverse chemical reaction 1 multiply with two:
Kc₂ = 1 ÷ (Kc₁)².
Kc₂ = 1 ÷ (0,020)².
Kc₂ = 1 ÷ 0,0004.
Kc₂ = 2500 = 2,5·10³.

Answer from: emalvidrez5205

Kc for reaction 2 at the same temperature

$\large{\boxed{\bold{Kc~=~50^2~=~2500}}}$

$\boxed{\boxed{\bold{Further~explanation}}}$

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

$\large{\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$

$Unit~Kc~is~M ^{(p + q) - (m + n)}$

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

$\large{\boxed{\bold{Kc~=~\frac{[C_2H_2][H_2]^3}{[CH_4]^2} }}}$

Then reversed

C₂H₂ (g) + 3H₂ (g) ⇄ 2CH₄ (g)

$\large{\boxed{\bold{Kc~=~\frac{[CH_4]^2}{[[C_2H_2][H_2]^3} }}}$

Kc = 1 / 0.02 = 50

Then reaction 2 :

2C₂H₂ (g) + 6H₂ (g) ⇄ 4CH₄ (g)

$\large{\boxed{\bold{Kc~=~\frac{[CH_4]^4}{[[C_2H_2]^2[H_2]^6} }}}$

$\large{\boxed{\bold{Kc~=~50^2~=~2500}}}$

$\boxed{\boxed{\bold{Learn~more}}}$

an equilibrium constant link

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Keywords: constant equilibrium, Kc, concentration, product, reactant, reaction coefficient

Given the information for reaction 1 at a specific temperature; calculate kc for reaction 2 at the

Answer from: Quest

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. $hbr$ completely dissociates to $h^+$ and $br-$ ions. a is incorrect because $h_2s$ does not fully ionize to $h^+$ and $hs^-$ ions and some remain as the compound $h_2s$ . c and d are incorrect because, $li_2o$ and $libr$ do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise, if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both $li_2o$ and $libr$ have a complete octet so can not accept any electrons which disqualifies them as acids.

Answer from: Quest

yes

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