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)

Answers: 1

Show answers

Answers

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.

answered: 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

link

Calculate the value of the equilibrium constant, Kc

link

Concentration of hi at equilibrium

link

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

answered: Guest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$